Recursion

http://www.zhitongguigu.com/

What is recursion

Recursion in computer science is a method where the solution to a problem depends on solutions to smaller instances of the same problem (as opposed to iteration).

http://www.zhitongguigu.com/

How to solve a problem

Divide into small problems.
Solve small problems.
Use the result of small problems to solve the original problem.
If the small problems are the same as original one, just scale is different. Then this is called recursion.

http://www.zhitongguigu.com/

Recursion Properties

Base case
- simple scenario that does not need recursion to produce an answer.
Recursion
- a set of rules that reduce all other case toward the base case.

http://www.zhitongguigu.com/

Mathematical Induction

To prove a given statement
- Base case, to prove the given statement for the first natural number.
- Inductive step, to prove that the given statement for any one natural number implies the given statement for the next natural number.

http://www.zhitongguigu.com/

Fibonacci Number

F_n = F_{n-1} + F_{n-2}

F_n = F_{n-1} + F_{n-2}

http://www.zhitongguigu.com/

Base Case: F(0) = 0;

F(1) = 1;

Recursion Rule: F(n) = F(n - 1) + F(n - 2)

Text

// Fibonacci calculation
int Fibonacci (int n) {
    //Base Case
    if(n == 0) return 0;
    if(n == 1) return 1;
    //recursion rule
    return Fibonacci(n - 1) + Fibonacci(n - 2);
}

                                   f(5)
=                    f(4)            +                     f(3)
=      f(3)            +    f(2)                f(2)         +  {f(1) = 1}
=  f(2) + {f(1) = 1}  {f(1)=1} + {f(0)=0}  {f(1)=1} + {f(0)=0}
={f(1)=1} + {f(0)=0}

Time Complexity

1 2 4 6 8 ... O(2^(n-1))

last layer nodes?

!!!binary search or tree problem BigO

What's recursion?

Just some special problem solving strategy

http://www.zhitongguigu.com/

Code problem with recursion

Base case
recursion rules
Represent the problem with coding function.

http://www.zhitongguigu.com/

How to solve a problem

Divide into small problems.
Solve small problems.
Use the result of small problems to solve the original problem.
If the small problems are the same as original one, just scale is different. Then this is called recursion.

function solveProblem() {

divide into small problems;

...

solveProblem();

}

Most programming languages support recursion by allowing functions to call itself within the program text.

http://www.zhitongguigu.com/

function tellStory() {

mountain();

temple();

oldMonk();

tellStory();

}

从前有个山，

山里有个庙，

庙里有个老和尚，

老和尚在给小和尚讲故事，

从前有个山...

讲故事

Old Monk Tells Story

http://www.zhitongguigu.com/

function factorial(int n) {

int m = factorial(n-1);

return m*n;

}

Factorial of n

calculate factorial of (n-1)，mark the result as m.
the factorial of n would be m*n

Factorial of n

http://www.zhitongguigu.com/

function factorial(int n) {

if (n==1) return 1;

int m = factorial(n-1);

return m*n;

}

Factorial of n

If n is 1, the factorial is 1.
calculate factorial of (n-1)，mark the result as m.
the factorial of n would be m*n

Factorial of n

function factorial(int n) {

if (n==1) return 1;

return m * factorial(n-1);

}

http://www.zhitongguigu.com/

function fibo(int n) {

if (n == 1) return 1;

if (n == 2) return 1;

return fibo(n-1) + fibo(n-2);

}

nth fibonacci number

If n is 1, then the number is 1.
If n is 2, then the number is 1.

The nth Fibonacci Number

F_n = F_{n-1} + F_{n-2}

F_n = F_{n-1} + F_{n-2}

1, 1, 2, 3, 5, 8, 13, 21, 34, ...

http://www.zhitongguigu.com/

Code problem with recursion

Base case
recursion rules
Represent the problem with coding function.
- Define the essential parameters
  - Parameters that define the problem.
- Define the return value

http://www.zhitongguigu.com/

Climb Building

1 stair left, one way.
2 stairs left, two ways.
n stairs left,
- climb one stair first, climb the (n-1) later.
- climb two stairs first, climb the (n-2) later.

There is a building with n stairs, one person can climb 1 or 2 stairs at one time, then how many different ways there are for this person to climb to the top.

public static int climb(int n) {

if (n <= 2) return n;

return climb(n-1) + climb(n-2);

}

http://www.zhitongguigu.com/

Code problem with recursion

Base case
recursion rules
Represent the problem with coding function.
- Define the essential parameters
  - Parameters that define the problem.
  - Parameters that store the temporary result or state.
- Define the return value

http://www.zhitongguigu.com/

Climb Building

There is a building with n stairs, one person can climb 1 or 2 stairs at one time. Print all the possible ways to climb to the top.

public static void climb(int n, String preWay) {
if (n == 1) {
System.out.println(preWay + " 1");
return;
}
if (n == 2) {
System.out.println(preWay + " 2");
System.out.println(preWay + " 1 1");
return;
}
String preWay1 = preWay + " 1";
climb(n-1, preWay1);
String preWay2 = preWay + " 2";
climb(n-2, preWay2);
}

http://www.zhitongguigu.com/

Towers of Hanoi

Move the first (n-1) disk from A to B
Move the nth disk from A to C
Move the other (n-1) disk from B to C

How many steps we need to move all n disks from A to C.

public static int hanoi(int n) {

if (n == 1) return 1;

return hanoi(n-1) + 1 + hanoi(n-1);

}

http://www.zhitongguigu.com/

Towers of Hanoi

Move the first (n-1) disk from A to B
Move the nth disk from A to C
Move the other (n-1) disk from B to C

How many steps we need to move all n disks from A to C.

public static int hanoi(int n) {

if (n == 1) return 1;

return 2 * hanoi(n-1) + 1;

}

http://www.zhitongguigu.com/

Towers of Hanoi

Move the first (n-1) disk from A to B
Move the nth disk from A to C
Move the other (n-1) disk from B to C

Print all the steps that is needed to move all n disks from A to C.

public static void hanoi(int n, char source, char spare, char target) {
if (n == 1) {
System.out.println("Move " + source + " to " + target);
return;
}
hanoi(n-1, source, target, spare);
System.out.println("Move " + source + " to " + target);
hanoi(n-1, spare, source, target);
}

http://www.zhitongguigu.com/

Towers of Hanoi

Move the first (n-1) disk from A to B
Move the nth disk from A to C
Move the other (n-1) disk from B to C

Print all the steps that is needed to move all n disks from A to C.

public static void hanoi(int n, char source, char spare, char target) {

if (n > 0) {

hanoi(n-1, source, target, spare);

System.out.println("Move " + source + " to " + target);

hanoi(n-1, spare, source, target);

}

http://www.zhitongguigu.com/

Sum of Linked List

If the list is empty, sum is zero.
Calculate the sum of linked list from head.next.
The final sum would be head.val and partial sum.

Calculate the sum of all nodes in a linked list.

http://www.zhitongguigu.com/

Sum of Linked List

If the list is empty, sum is zero.
Calculate the sum of linked list from head.next.
The final sum would be head.val and partial sum.

Calculate the sum of all nodes in a linked list.

public static int sum(ListNode head) {

if (head == null) return 0;

return head.val + sum(head.next);

}

http://www.zhitongguigu.com/

Remove Linked List Elements

If the head is null, nothing needs to be done.
Remove elements in linked list from head.next, store as newHead.
If value of head node is val, then remove head.
- Otherwise link the newHead to head.

Remove all elements from a linked list of integers that have value val.

public static ListNode removeElements(ListNode head, int val) {

if (head == null) return null;

ListNode newHead = removeElements(head.next, val);

if (head.val == val) {

return newHead;

} else {

head.next = newHead;

return head;

}

http://www.zhitongguigu.com/

Problem

sub-problem-1

sub-sub-problem

sub-sub-sub-problem

base-problem

Problem

sub-problem-1

base-problem

sub-problem-2

sub-problem-3

sub-sub-problem-2

sub-sub-sub-problem

sub-sub-problem-1

...

http://www.zhitongguigu.com/

Problem

sub-problem-1

sub-sub-problem

sub-sub-sub-problem

base-problem

Problem

sub-problem-1

base-problem

sub-problem-2

sub-problem-3

sub-sub-problem-2

sub-sub-sub-problem

sub-sub-problem-1

...

http://www.zhitongguigu.com/

Problem

sub-problem-1

sub-sub-problem

sub-sub-sub-problem

base-problem

Problem

sub-problem-1

base-problem

sub-problem-2

sub-problem-3

sub-sub-problem-2

sub-sub-sub-problem

sub-sub-problem-1

...

http://www.zhitongguigu.com/

Problem

sub-problem-1

sub-sub-problem

sub-sub-sub-problem

base-problem

Problem

sub-problem-1

base-problem

sub-problem-2

sub-problem-3

sub-sub-problem-2

sub-sub-sub-problem

sub-sub-problem-1

...

http://www.zhitongguigu.com/

Problem

sub-problem-1

sub-sub-problem

sub-sub-sub-problem

base-problem

Problem

sub-problem-1

base-problem

sub-problem-2

sub-problem-3

sub-sub-problem-2

sub-sub-sub-problem

sub-sub-problem-1

...

http://www.zhitongguigu.com/

Problem

sub-problem-1

sub-sub-problem

sub-sub-sub-problem

base-problem

Problem

sub-problem-1

sub-sub-problem

sub-sub-sub-problem

base-problem

Problem

sub-problem-1

base-problem

sub-problem-2

sub-problem-3

sub-sub-problem-2

sub-sub-sub-problem

sub-sub-problem-1

...

Backtracking

http://www.zhitongguigu.com/

0-1 Knapsack

Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Return whether we can pick specific items so that their total weight s.

Example Input:

s = 20;

w = [14, 8, 7, 5, 3];

Example Output:

True;

http://www.zhitongguigu.com/

Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Return whether we can pick specific items so that their total weight s.

Example Input:

s = 20;

w = [14, 8, 7, 5, 3];

Example Output:

True;

Given an item, just two options:

pick it, the problem become (s-w[i], w - w[i])
not pick it, the problem become (s, w - w[i])

http://www.zhitongguigu.com/

0-1 Knapsack

Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Return whether we can pick specific items so that their total weight s.

public static boolean knapsack(int s, int[] weights, int index) {
    if (s == 0) {
      return true;
    }
    if (s < 0 || index >= weights.length) {
      return false;
    }
    return knapsack(s - weights[index], weights, index+1) ||
      knapsack(s, weights, index+1);
}

http://www.zhitongguigu.com/

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

Example Input:

Start Point: (0, 0); Target Point (5, 5);

Maze: char[][] = {

{'.', 'X', '.', '.', '.', 'X'},

{'.', '.', '.', 'X', '.', 'X'},

{'X', 'X', '.', 'X', '.', '.'},

{'.', 'X', 'X', 'X', '.', 'X'},

{'.', '.', '.', '.', '.', 'X'},

{'.', '.', '.', '.', '.', '.'}

}

Example Output: True

http://www.zhitongguigu.com/

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

Example Input:

Start Point: (0, 0); Target Point (5, 5);

Maze: char[][] = {

{'.', 'X', '.', '.', '.', 'X'},

{'.', '.', '.', 'X', '.', 'X'},

{'X', 'X', '.', 'X', '.', '.'},

{'.', 'X', 'X', 'X', '.', 'X'},

{'.', '.', '.', '.', '.', 'X'},

{'.', '.', '.', '.', '.', '.'}

}

Example Output: True

http://www.zhitongguigu.com/

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

http://www.zhitongguigu.com/

Out of Bound
Wall

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

http://www.zhitongguigu.com/

Out of Bound
Wall
Visited
- (1, 2) -> (2, 2) -> (1, 2) -> (2, 2) -> ...

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

http://www.zhitongguigu.com/

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

http://www.zhitongguigu.com/

public static boolean solveMaze(char[][] maze, 
                                int startX, int startY, int targetX, int targetY,
                                boolean[][] visited) {
    if (startX < 0 || startX >= maze.length ||
        startY < 0 || startY >= maze[0].length ||
        maze[startX][startY] == 'X' || visited[startX][startY]) {
      return false;
    }
    if (startX == targetX && startY == targetY) {
      return true;
    }
    visited[startX][startY] = true;
    if (solveMaze(maze, startX + 1, startY, targetX, targetY, visited) ||
        solveMaze(maze, startX, startY + 1, targetX, targetY, visited) ||
        solveMaze(maze, startX - 1, startY, targetX, targetY, visited) ||
        solveMaze(maze, startX, startY - 1, targetX, targetY, visited)) {
      return true;
    }
    return false;
}

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

http://www.zhitongguigu.com/

public static boolean solveMaze(char[][] maze, 
                                int startX, int startY, int targetX, int targetY) {

    if (startX < 0 || startX >= maze.length ||
        startY < 0 || startY >= maze[0].length ||
        maze[startX][startY] == 'X') {
      return false;
    }
    if (startX == targetX && startY == targetY) {
      return true;
    }
    maze[startX][startY] = 'X';
    if (solveMaze(maze, startX + 1, startY, targetX, targetY) ||
        solveMaze(maze, startX, startY + 1, targetX, targetY) ||
        solveMaze(maze, startX - 1, startY, targetX, targetY) ||
        solveMaze(maze, startX, startY - 1, targetX, targetY)) {
      return true;
    }
    return false;
}

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

http://www.zhitongguigu.com/

public static boolean solveMaze(char[][] maze, 
                                int startX, int startY, int targetX, int targetY) {

    if (startX < 0 || startX >= maze.length ||
        startY < 0 || startY >= maze[0].length ||
        maze[startX][startY] == 'X') {
      return false;
    }
    if (startX == targetX && startY == targetY) {
      return true;
    }
    maze[startX][startY] = 'X';
    int[] dx = {1, 0, -1, 0};
    int[] dy = {0, 1, 0, -1};
    for (int i = 0; i < 4; i++) {
      if (solveMaze(maze, startX + dx[i], startY + dy[i], targetX, targetY)) {
        return true;
      }
    }
    return false;
}

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

http://www.zhitongguigu.com/

public static boolean solveMaze(char[][] maze, 
                                int startX, int startY, int targetX, int targetY) {

    if (startX == targetX && startY == targetY) {
      return true;
    }
    maze[startX][startY] = 'X';
    int[] dx = {1, 0, -1, 0};
    int[] dy = {0, 1, 0, -1};
    for (int i = 0; i < 4; i++) {
      int newX = startX + dx[i], newY = startY + dy[i];
      if (newX < 0 || newX >= maze.length || newY < 0 || newY >= maze.length ||
          maze[newX][newY] == 'X') {
        continue;
      }
      if (solveMaze(maze, newX, newY, targetX, targetY)) {
        return true;
      }
    }
    return false;
}

Maze

Given a maze and a start point and a target point, return whether the target can be reached.

http://www.zhitongguigu.com/

public static boolean solveMaze(char[][] maze, 
                                int startX, int startY, int targetX, int targetY,
                                String path) {
    if (startX < 0 || startX >= maze.length ||
        startY < 0 || startY >= maze[0].length ||
        maze[startX][startY] == 'X') {
      return false;
    }
    if (startX == targetX && startY == targetY) {
      System.out.println(path);
      return true;
    }
    maze[startX][startY] = 'X';
    int[] dx = {1, 0, -1, 0};
    int[] dy = {0, 1, 0, -1};
    char[] direction = {'D', 'R', 'U', 'L'};
    for (int i = 0; i < 4; i++) {
      String newPath = path + direction[i] + " ";
      if (solveMaze(maze, startX+dx[i], startY+dy[i], targetX, targetY, newPath)) {
        return true;
      }
    }
    return false;
}

Maze

Given a maze and a start point and a target point, print out the path to reach the target.

http://www.zhitongguigu.com/

public static boolean solveMaze(char[][] maze, 
                                int startX, int startY, int targetX, int targetY,
                                ArrayList<Character> path) {
    if (startX < 0 || startX >= maze.length ||
        startY < 0 || startY >= maze[0].length ||
        maze[startX][startY] == 'X') {
      return false;
    }
    if (startX == targetX && startY == targetY) {
      return true;
    }
    maze[startX][startY] = 'X';
    int[] dx = {1, 0, -1, 0};
    int[] dy = {0, 1, 0, -1};
    char[] direction = {'D', 'R', 'U', 'L'};
    for (int i = 0; i < 4; i++) {
      path.add(direction[i]);
      if (solveMaze(maze, startX+dx[i], startY+dy[i], targetX, targetY, path)) {
        return true;
      }
      path.remove(path.size()-1);
    }
    return false;
}

Maze

Given a maze and a start point and a target point, return the path to reach the target.

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

..Q.....

.....Q..

...Q....

.Q......

.......Q

....Q...

......Q.

Q.......

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

Q	X	X	X	X	X	X	X
X	X		X			X
X		X		X		X
X			X		X	X	X
X	X	X	X	X	X	Q	X
X					X	X	X
X				X		X
X			X			X	X

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

X	Q	X	X	X	X	X	X
X	X	X
	X		X
	X			X
	X				X
	X					X
	X						X
	X

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

X	Q	X	X	X	X	X	X
X	X	X	Q	X	X	X	X
	X	X	X	X
	X		X	X	X
X	X		X		X	X	X
	X		X			X
	X		X				X
	X		X

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

X	Q	X	X	X	X	X	X
X	X	X	Q	X	X	X	X
X	X	X	X	X	Q	X	X
	X		X	X	X	X
X	X		X		X	X	X
	X	X	X		X	X
X	X		X		X		X
X	X		X		X

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

X	Q	X	X	X	X	X	X
X	X	X	Q	X	X	X	X
X	X	X	X	X	Q	X	X
X	X	X	X	X	X	X	Q
X	X		X		X	X	X
	X	X	X		X	X	X
X	X		X	X	X		X
X	X		X		X		X

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

X	Q	X	X	X	X	X	X
X	X	X	Q	X	X	X	X
X	X	X	X	X	Q	X	X
X	X	X	X	X	X	X	Q
X	X	Q	X	X	X	X	X
	X	X	X		X	X	X
X	X	X	X	X	X		X
X	X	X	X		X		X

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

X	Q	X	X	X	X	X	X
X	X	X	Q	X	X	X	X
X	X	X	X	X	Q	X	X
X	X	X	X	X	X	X	Q
X	X	Q	X	X	X	X	X
Q	X	X	X	X	X	X	X
X	X	X	X	X	X		X
X	X	X	X		X		X

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

X	Q	X	X	X	X	X	X
X	X	X	Q	X	X	X	X
X	X	X	X	X	Q	X	X
X	X	X	X	X	X	X	Q
X	X	Q	X	X	X	X	X
Q	X	X	X	X	X	X	X
X	X	X	X	X	X	Q	X
X	X	X	X		X	X	X

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

X	Q	X	X	X	X	X	X
X	X	X	Q	X	X	X	X
X	X	X	X	X	Q	X	X
X	X	X	X	X	X	X	Q
X	X	Q	X	X	X	X	X
Q	X	X	X	X	X	X	X
X	X	X	X	X	X	Q	X
X	X	X	X	Q	X	X	X

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

void putQueens(char[][] board, int row,
               List<List<String>> results) {

    if (row == board.length) {
        saveResult(board, results);
        return;
    }
    for (int col = 0; col < board[0].length; col++) {
        if (isLegal(board, row, col)) {
            board[row][col] = 'Q';
            putQueens(board, row+1, results);
            board[row][col] = '.';
        }
    }
    return;
}

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

void saveResult(char[][] board,
                List<List<String>> results) {
    List<String> result = new ArrayList<>();
    for (int i = 0; i < board.length; i++) {
        result.add(new String(board[i]));
    }
    results.add(result);
}

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

boolean isLegal(char[][] board, int row, int col) {
    for (int i = row-1; i >= 0; i--) {
        if (board[i][col] == 'Q')
            return false;
            
        int tempCol = col - row + i;
        if (tempCol >= 0 &&
            tempCol < board.length &&
            board[i][tempCol] == 'Q')
            return false;
            
        tempCol = col + row - i;
        if (tempCol >= 0 &&
            tempCol < board.length &&
            board[i][tempCol] == 'Q')
            return false;
    }
    return true;
}

http://www.zhitongguigu.com/

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

public List<List<String>> solveNQueens(int n) {
    List<List<String>> results = new ArrayList<>();
    char[][] board = new char[n][n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            board[i][j] = '.';
        }
    }
    putQueens(board, 0, results);
    return results;
}

http://www.zhitongguigu.com/

public List<List<String>> solveNQueens(int n) {
    List<List<String>> results = new ArrayList<>();
    char[][] board = new char[n][n];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            board[i][j] = '.';
        }
    }
    putQueens(board, 0, results);
    return results;
}

void putQueens(char[][] board, int row, List<List<String>> results) {
    if (row == board.length) {
        saveResult(board, results);
        return;
    }
    for (int col = 0; col < board[0].length; col++) {
        if (isLegal(board, row, col)) {
            board[row][col] = 'Q';
            putQueens(board, row+1, results);
            board[row][col] = '.';
        }
    }
    return;
}

void saveResult(char[][] board, List<List<String>> results) {
    List<String> result = new ArrayList<>();
    for (int i = 0; i < board.length; i++) {
        result.add(new String(board[i]));
    }
    results.add(result);
}

boolean isLegal(char[][] board, int row, int col) {
    for (int i = row-1; i >= 0; i--) {
        if (board[i][col] == 'Q')
            return false;
            
        int tempCol = col - row + i;
        if (tempCol >= 0 && tempCol < board.length && board[i][tempCol] == 'Q')
            return false;
            
        tempCol = col + row - i;
        if (tempCol >= 0 && tempCol < board.length && board[i][tempCol] == 'Q')
            return false;
    }
    return true;
}

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

We don't need a char[][] to store the board
- int[] will be enough

http://www.zhitongguigu.com/

public List<List<String>> solveNQueens(int n) {
    List<List<String>> results = new ArrayList<>();
    int[] board = new int[n];
    putQueens(board, 0, results);
    return results;
}

void putQueens(int[] board, int row, List<List<String>> results) {
    if (row == board.length) {
        saveResult(board, results);
        return;
    }
    for (int col = 0; col < board.length; col++) {
        if (isLegal(board, row, col)) {
            board[row] = col;
            putQueens(board, row+1, results);
        }
    }
    return;
}

boolean isLegal(int[] board, int row, int col) {
    for (int i = row-1; i >= 0; i--) {
        if (board[i] == col || Math.abs(board[i] - col) == Math.abs(row - i)) {
            return false;
        }
    }
    return true;
}

void saveResult(int[] board, List<List<String>> results) {
    List<String> result = new ArrayList<>();
    for (int i = 0; i < board.length; i++) {
        StringBuilder oneLine = new StringBuilder();
        for (int j = 0; j < board.length; j++) {
            if (j == board[i]) {
                oneLine.append('Q');
            } else {
                oneLine.append('.');
            }
        }
        result.add(oneLine.toString());
    }
    results.add(result);
}

Eight Queens

The eight queens puzzle is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other.

public int totalNQueens(int n) {
    int upper = (1 << n) -1;
    int[] count = {0};
    queen(0, 0, 0, upper, count);
    return count[0];
}

public void queen(int row, int ld, int rd, int upper, int[] count) {
    if (row != upper) {
        int pos = upper & (~(row | ld | rd));
        while (pos != 0) {
            int p = pos & (-pos);
            pos -= p;
            queen(row+p, (ld+p) << 1, (rd+p) >> 1, upper, count);
        }
    } else {
        count[0]++;
    }
}

http://www.zhitongguigu.com/

Backtracking Summary

http://www.zhitongguigu.com/

Backtrack = try, iterate, traverse, etc.
Keep trying (in the search space) until
- Solution is found
- No more meaningful methods to try (no more search space)
Level-N problem -> M * Level-(N-1) subproblem
- Keep states the same when entering subproblem except shared fields.

Knapsack

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

All candidate numbers are unique.

The same repeated number may be chosen from C unlimited number of times.

Note:

All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.

Example Input: [7], [2, 3, 6, 7]

Example Output: [[2, 2, 3], [7]]

http://www.zhitongguigu.com/

Knapsack

http://www.zhitongguigu.com/

Define the problem as (C, i, T), then given a number

Pick => (C, i, T-C[i])
Not Pick => (C, i+1, T-C[i])

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

All candidate numbers are unique.

The same repeated number may be chosen from C unlimited number of times.

Knapsack

http://www.zhitongguigu.com/

public static ArrayList<ArrayList<Integer>> knapsack(int[] candidates, 
                                                     int target) {
    Arrays.sort(candidates);
    ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
    ArrayList<Integer> cur = new ArrayList<Integer>();
    knapsack(candidates, 0, target, results, cur);
    return results;
}

public static void knapsack(int[] candidates, int index, int target,
                            ArrayList<ArrayList<Integer>> results,
                            ArrayList<Integer> cur) {
    if (target < 0 || (target != 0 && index == candidates.length)) {
        return;
    }
    if (target == 0) {
        results.add(new ArrayList<Integer>(cur));
        return;
    }
    cur.add(candidates[index]);
    knapsack(candidates, index, target-candidates[index], results, cur);
    cur.remove(cur.size()-1);
    knapsack(candidates, index+1, target, results, cur);
}

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Knapsack

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

All candidate numbers are unique.

The same repeated number may be chosen from C unlimited number of times.

http://www.zhitongguigu.com/

Iterate all numbers to decide whether to pick the current number.

<=>

Which one of the numbers should I pick as the smallest one in the current set.

Knapsack

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

public ArrayList<ArrayList<Integer>> knapsack(int[] candidates, int target) {
    Arrays.sort(candidates);
    ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
    ArrayList<Integer> cur = new ArrayList<Integer>();
    knapsack(candidates, 0, target, results, cur);
    return results;
}

public void knapsack(int[] candidates, int index, int target,
                     ArrayList<ArrayList<Integer>> results,
                     ArrayList<Integer> cur) {
    if (target < 0) {
        return;
    }
    if (target == 0) {
        results.add(new ArrayList<Integer>(cur));
        return;
    }
    for (int i = index; i < candidates.length; i++) {
        cur.add(candidates[i]);
        knapsack(candidates, i, target-candidates[i], results, cur);
        cur.remove(cur.size()-1);
    }
    return;
}

http://www.zhitongguigu.com/

Knapsack II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Candidate numbers may contain duplicate.

Each number in C may only be used once in the combination.

public List<List<Integer>> knapsack(int[] candidates, int target) {
    Arrays.sort(candidates);
    List<List<Integer>> results = new ArrayList<List<Integer>>();
    List<Integer> cur = new ArrayList<Integer>();
    knapsack(candidates, 0, target, results, cur);
    return results;
}

public void knapsack(int[] candidates, int index, int target,
                     List<List<Integer>> results, List<Integer> cur) {
    if (target < 0)  return;
    if (target == 0) {
        results.add(new ArrayList<Integer>(cur));
        return;
    }
    for (int i = index; i < candidates.length; i++) {
        cur.add(candidates[i]);
        knapsack(candidates, i+1, target-candidates[i], results, cur);
        cur.remove(cur.size()-1);
        while (i < candidates.length-1 && candidates[i] == candidates[i+1]) i++;
    }
    return;
}

http://www.zhitongguigu.com/

0-1 Knapsack II

Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Try to put items into the pack as many as possible, return the largest weight we can get in the knapsack.

public static int knapsack(int s, int[] weights, int index) {
    if (s == 0 || index == weights.length) {
      return 0;
    }
    if (weights[index] > s) {
      return knapsack(s, weights, index+1);
    }
    return Math.max(knapsack(s, weights, index+1),
                    weights[index] + knapsack(s-weights[index], weights, index+1));
}

http://www.zhitongguigu.com/

Generate Parenthesis

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example:

Given n = 3,

return ["((()))", "(()())", "(())()", "()(())", "()()()" ]

http://www.zhitongguigu.com/

Generate Parenthesis

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example:

Given n = 3,

return ["((()))", "(()())", "(())()", "()(())", "()()()" ]

http://www.zhitongguigu.com/

n parenthesis => n left + n right => 2*n positions.
Each time, one left or right should be chosen.
- We can always pick left.
- If there are more right remained, we can put right.

http://www.zhitongguigu.com/

Generate Parenthesis

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

http://www.zhitongguigu.com/

public ArrayList<String> generateParenthesis(int n) {
    ArrayList<String> results = new ArrayList<>();
    generateParenthesis(results, "", n, n);
    return results;
}

public void generateParenthesis(ArrayList<String> results, String prefix,
                                int left, int right) {
    if (left == 0 && right == 0) {
        results.add(prefix);
        return;
    }
    if (left > 0) {
        generateParenthesis(results, prefix+"(", left-1, right);
    }
    if (left < right) {
        generateParenthesis(results, prefix+")", left, right-1);
    }
}

Permutation (Leetcode 46)

Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

public List<List<Integer>> permute(int[] nums) {
// Implement this method.
}

http://www.zhitongguigu.com/

Given a collection of distinct numbers, return all possible permutations.

public ArrayList<ArrayList<Integer>> permute(int[] num) {
    ArrayList<Integer> numList = new ArrayList<Integer>();
    Arrays.sort(num);
    for (int i = 0; i < num.length; i++)
        numList.add(num[i]);
    return permute(new ArrayList<Integer>(), numList);
}

public ArrayList<ArrayList<Integer>> permute(ArrayList<Integer> cur,
                                             ArrayList<Integer> num) {
    ArrayList<ArrayList<Integer>> results =
         new ArrayList<ArrayList<Integer>>();
    if (num.size() == 0) {
        results.add(cur);
        return results;
    }
    for (int i = 0; i < num.size(); i++) {
        ArrayList<Integer> newCur = new ArrayList<Integer>(cur);
        newCur.add(num.get(i));
        ArrayList<Integer> newNum = new ArrayList<Integer>(num);
        newNum.remove(i);
        results.addAll(permute(newCur, newNum));
    }
    return results;
}

http://www.zhitongguigu.com/

Permutation (Leetcode 46)

Given a collection of distinct numbers, return all possible permutations.

public List<List<Integer>> permute(int[] num) {
    Arrays.sort(num);
    return permute(new ArrayList<Integer>(), num);
}

public List<List<Integer>> permute(ArrayList<Integer> cur, int[] num) {
    List<List<Integer>> results = new ArrayList<List<Integer>>();
    if (cur.size() == num.length) {
        results.add(new ArrayList<Integer>(cur));
        return results;
    }
    for (int i = 0; i < num.length; i++) {
        if (cur.contains(num[i])) {
            continue;
        }
        cur.add(num[i]);
        results.addAll(permute(cur, num));
        cur.remove(cur.size() - 1);
    }
    return results;
}

http://www.zhitongguigu.com/

Permutation (Leetcode 46)

Given a collection of distinct numbers, return all possible permutations.

public List<List<Integer>> permute(int[] num) {
    List<List<Integer>> results = new ArrayList<List<Integer>>();
    Arrays.sort(num);
    permute(results, new ArrayList<Integer>(), num);
    return results;
}
    
public void permute(List<List<Integer>> results,
                    ArrayList<Integer> cur, int[] num) {
    if (cur.size() == num.length) {
        results.add(new ArrayList<Integer>(cur));
        return;
    }
    for (int i = 0; i < num.length; i++) {
        if (cur.contains(num[i])) {
            continue;
        }
        cur.add(num[i]);
        permute(results, cur, num);
        cur.remove(cur.size() - 1);
    }
}

http://www.zhitongguigu.com/

Permutation (Leetcode 46)

Given a collection of distinct numbers, return all possible permutations.

public List<List<Integer>> permute(int[] num) {
    List<List<Integer>> results = new ArrayList<List<Integer>>();
    Arrays.sort(num);
    permute(results, num, 0);
    return results;
}

public void permute(List<List<Integer>> results, int[] num, int index) {
    if (index == num.length) {
	ArrayList<Integer> result = new ArrayList<>();
	for (int i = 0; i < num.length; i++) {
	    result.add(num[i]);
	}
	results.add(result);
	return;
    }
    for (int i = index; i < num.length; i++) {
	swap(num, index, i);
	permute(results, num, index + 1);
	swap(num, index, i);
    }
}

public void swap(int[] num, int i, int j) {
    int temp = num[i];
    num[i] = num[j];
    num[j] = temp;
}

http://www.zhitongguigu.com/

Permutation II (Leetcode 47)

Given a collection of ~~distinct~~ numbers, return all possible permutations.

public List<List<Integer>> permuteUnique(int[] num) {
    ArrayList<Integer> numList = new ArrayList<>();
    Arrays.sort(num);
    for (int i = 0; i < num.length; i++)
        numList.add(num[i]);
    return permute(new ArrayList<Integer>(), numList);
}

public List<List<Integer>> permute(
    ArrayList<Integer> cur, ArrayList<Integer> num) {

    List<List<Integer>> results = new ArrayList<>();
    if (num.size() == 0) {
        results.add(cur);
        return results;
    }
    for (int i = 0; i < num.size(); i++) {
        if (i > 0 && num.get(i) == num.get(i-1))
            continue;
        ArrayList<Integer> newCur = new ArrayList<>(cur);
        newCur.add(num.get(i));
        ArrayList<Integer> newNum = new ArrayList<>(num);
        newNum.remove(i);
        results.addAll(permute(newCur, newNum));
    }
    return results;
}

http://www.zhitongguigu.com/

Permutation II (Leetcode 47)

Given a collection of ~~distinct~~ numbers, return all possible permutations.

public List<List<Integer>> permuteUnique(int[] num) {
    Arrays.sort(num);
    boolean[] visited = new boolean[num.length];
    for (int i = 0; i < visited.length; i++) {
        visited[i] = false;
    }
    return permute(new ArrayList<Integer>(), num, visited);
}

public List<List<Integer>> permute(ArrayList<Integer> cur, int[] num, boolean[] visited) {
    List<List<Integer>> results = new ArrayList<>();
    if (cur.size() == num.length) {
        results.add(new ArrayList<Integer>(cur));
        return results;
    }
    for (int i = 0; i < num.length; i++) {
        if (visited[i] || (i > 0 && num[i] == num[i-1] && !visited[i-1])) {
            continue;
        }
        visited[i] = true;
        cur.add(num[i]);
        results.addAll(permute(cur, num, visited));
        cur.remove(cur.size() - 1);
        visited[i] = false;
    }
    return results;
}

http://www.zhitongguigu.com/

Combination

Given a collection of distinct numbers, return all possible combinations.

For example,
[2, 6, 8] have the following permutations:
[], [2], [6], [8], [2, 6], [2, 8], [6, 8], [2, 6, 8].

public List<List<Integer>> combine(int[] nums) {
// Implement this method.
}

http://www.zhitongguigu.com/

Combination

Given a collection of distinct numbers, return all possible combinations.

public static List<List<Integer>> combine(int[] nums) {
    List<List<Integer>> results = new ArrayList<>();
    combination(results, nums, 0, new ArrayList<Integer>());
    return results;
}

public static void combination(List<List<Integer>> results, 
                               int[] nums, int index, ArrayList<Integer> items) {
    if (index == nums.length) {
      results.add(items);
      return;
    }
    ArrayList<Integer> newItems1 = new ArrayList<Integer>(items);
    combination(results, nums, index+1, newItems1);
    
    ArrayList<Integer> newItems2 = new ArrayList<Integer>(items);
    newItems2.add(nums[index]);
    combination(results, nums, index+1, newItems2);
  }

http://www.zhitongguigu.com/

Combination

Given a collection of distinct numbers, return all possible combinations.

public static List<List<Integer>> combine(int[] nums) {
    List<List<Integer>> results = new ArrayList<>();
    combination(results, nums, 0, new ArrayList<Integer>());
    return results;
}

public static void combination(List<List<Integer>> results, 
                               int[] nums, int index, ArrayList<Integer> items) {
    if (index == nums.length) {
      results.add(new ArrayList<Integer>(items));
      return;
    }
    combination(results, nums, index+1, items);
    
    items.add(nums[index]);
    combination(results, nums, index+1, items);
    items.remove(items.size()-1);
  }

http://www.zhitongguigu.com/

Combination

Given a collection of distinct numbers, return all possible combinations.

public static List<List<Integer>> combine(int[] nums) {
    List<List<Integer>> results = new ArrayList<>();
    combination(results, nums, 0, new ArrayList<Integer>());
    return results;
}

public static void combination(List<List<Integer>> results, 
                               int[] nums, int index,
                               ArrayList<Integer> items) {
    if (index == nums.length) {
      results.add(new ArrayList<Integer>(items));
      return;
    }
    for (int i = index; i < nums.length; i++) {
      items.add(nums[i]);
      combination(results, nums, i+1, items);
      items.remove(items.size()-1);
    }
}

http://www.zhitongguigu.com/

The empty set is missing.

-There will be at least one element in the "items"

Combination

Given a collection of distinct numbers, return all possible combinations.

public static List<List<Integer>> combine(int[] nums) {
    List<List<Integer>> results = new ArrayList<>();
    combination(results, nums, 0, new ArrayList<Integer>());
    return results;
}

public static void combination(List<List<Integer>> results, 
                               int[] nums, int index,
                               ArrayList<Integer> items) {
    if (index == nums.length) {
      results.add(new ArrayList<Integer>(items));
      return;
    }
    for (int i = index; i <= nums.length; i++) {
      if (i == nums.length) {
        combination(results, nums, i, items);
        return;
      }
      items.add(nums[i]);
      combination(results, nums, i+1, items);
      items.remove(items.size()-1);
    }
}

http://www.zhitongguigu.com/

This adds an empty set.

Combination

Given a collection of distinct numbers, return all possible combinations.

public static List<List<Integer>> combine(int[] nums) {
    List<List<Integer>> results = new ArrayList<>();
    combination(results, nums, 0, new ArrayList<Integer>());
    return results;
}

public static void combination(List<List<Integer>> results, 
                               int[] nums, int index,
                               ArrayList<Integer> items) {
    if (index == nums.length) {
      results.add(new ArrayList<Integer>(items));
      return;
    }
    for (int i = index; i < nums.length; i++) {
      items.add(nums[i]);
      combination(results, nums, i+1, items);
      items.remove(items.size()-1);
    }
    combination(results, nums, nums.length, items);
}

http://www.zhitongguigu.com/

This is the same as 89.4

Summary

http://www.zhitongguigu.com/

Recursion

subproblem

-> problem

subproblem-1 ||

subproblem-2 ||

subproblem-N

-> problem

subproblem-1 &&

subproblem-2 &&

subproblem-N

-> problem

Factorial,

Sum of LinkedList,

Remove Linked List Element, etc.

Maze,

Sudoku,

Most DFS problems, etc.

Knapsack

Permutations,

Combinations,

Eight Queen,

kSum, etc.

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

Pick one number from the array and do "partition sort"
- The left part will always be smaller than pivot
- The right part will always be larger than pivot
Quicksort the left part
Quicksort the right part

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

5	7	3	2	6

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

	7	3	2	6

pivot = 5

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

	7	3	2	6

pivot = 5

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

2	7	3		6

pivot = 5

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

2	7	3		6

pivot = 5

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

2		3	7	6

pivot = 5

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

2		3	7	6

pivot = 5

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

2	3		7	6

pivot = 5

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

2	3		7	6

pivot = 5

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

2	3	5	7	6

pivot = 5

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

Recursion & Two Pointers

http://www.zhitongguigu.com/

Quicksort an array

public static void quicksort(int[] nums, int begin, int end) {
    if (begin >= end) {
        return;
    }
    int pivotPostion = partition(nums, begin, end);
    quicksort(nums, begin, pivotPostion - 1);
    quicksort(nums, pivotPostion + 1, end);
}
  
public static int partition(int[] nums, int begin, int end) {
    int pivot = nums[begin];
    while (begin < end) {
        while (begin < end && nums[end] >= pivot) {
            end--;
        }
        nums[begin] = nums[end];
        while (begin < end && nums[begin] <= pivot) {
            begin++;
        }
        nums[end] = nums[begin];
    }
    nums[begin] = pivot;
    return begin;
}

Recursion & LinkedList

http://www.zhitongguigu.com/

Reverse a linked list

Reverse head.next, mark as newHead
newTail.next = head
return newHead;

Recursion & LinkedList

http://www.zhitongguigu.com/

Reverse a linked list

Reverse head.next, mark as newHead
newTail.next = head
return newHead;

Recursion & LinkedList

http://www.zhitongguigu.com/

Reverse a linked list

Reverse head.next, mark as newHead
newTail.next = head
return newHead;

Recursion & LinkedList

http://www.zhitongguigu.com/

Reverse a linked list

Reverse head.next, mark as newHead
newTail.next = head
return newHead;

Recursion & LinkedList

http://www.zhitongguigu.com/

Reverse a linked list

Reverse head.next, mark as newHead
newTail.next = head
return newHead;

public ListNode reverseList(ListNode head) {
    if (head == null || head.next == null) {
        return head;
    }
    ListNode newHead = reverseList(head.next);
    head.next.next = head;
    head.next = null;
    return newHead;
}

http://www.zhitongguigu.com/

Summary

Recursion is a strategy.
Always try to split a problem into sub-problems and solve the sub-problem first.
- If the solution is the same, then you can call the same method.

Homework

http://www.zhitongguigu.com/

Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Try to put items into the pack as many as possible, print out all the items that can get the largest weight. Each item can only get picked once.

public int[] knapsack(int s, int[] weights) {

// Please implement this method.

}

vector<int> knapsack(int s, vector<int> weights {

// Please implement this method.

}

0-1 Knapsack (Required)

http://www.zhitongguigu.com/

Homework (Required)

http://www.zhitongguigu.com/

Letter Combinations of a Phone Number

Homework (Optional)

http://www.zhitongguigu.com/

Subsets II

Sudoku Solver

Palindrome Partitioning

Restore IP Address

[GoValley-201612] Recursion

By govalley201612

[GoValley-201612] Recursion

Recursion

What is recursion

Recursion in computer science is a method where the solution to a problem depends on solutions to smaller instances of the same problem (as opposed to iteration).

How to solve a problem

Recursion Properties

Mathematical Induction

Time Complexity

1 2 4 6 8 ... O(2^(n-1))

More recursion problems

Greatest Common Divisor

Binary Search

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Recursive Definition

What's recursion?

Just some special problem solving strategy

Code problem with recursion

How to solve a problem

Old Monk Tells Story

Factorial of n

Factorial of n

The nth Fibonacci Number

Code problem with recursion

Climb Building

Code problem with recursion

Climb Building

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Towers of Hanoi

Sum of Linked List

Sum of Linked List

Remove Linked List Elements

0-1 Knapsack

0-1 Knapsack

0-1 Knapsack

0-1 Knapsack

0-1 Knapsack

0-1 Knapsack

0-1 Knapsack

0-1 Knapsack

0-1 Knapsack

0-1 Knapsack

0-1 Knapsack

Maze

Maze

Maze

Maze

Maze

Maze

Maze

Maze

Maze

Maze

Maze

Maze

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens

Eight Queens