Stack & Queue

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Stack & Queue

Two Different Data Structure
Stack: LIFO (Last in first out)
Queue: FIFO (First in first out)

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Stack: LIFO

LIFO: Last In First Out

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Tower of Hanoi

Queue: FIFO

FIFO: First In First Out

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Queue: FIFO

FIFO: First In First Out

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Stack

Stack<String> stack = new Stack<String>();
Stack<String> stack = new Stack<>();

Java

C++

stack <string> cards;

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Stack<String> stack = new Stack<String>();
Stack<String> stack = new Stack<>();

List<Map<Integer,Set<String>>> p = new ArrayList<Map<Integer,Set<String>>>();
List<Map<Integer,Set<String>>> p = new ArrayList<>();

// polymorphism
List<Map<Integer,Set<String>>> p = new ArrayList<>();
ArrayList<Map<Integer,Set<String>>> p = new ArrayList<>();

Java Diamond Operator

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Basic Operation

size
pop
push
peek

size
pop
push
top

Java

C++

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Stack

How to use Array to implement a stack?

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🤔

What might be the problems

Array -> Stack

class Stack<T> {
  private int top;
  int capacity;
  T[] stack;
  public Stack(int inCapacity) {
    capacity = inCapacity;
    stack = (T[]) new Object[capacity];
    top = -1;
  }
  public int size() {
    return top + 1;
  }
  public void push(T value) {
    if(top + 1 == capacity) {
      throw new IllegalStateException("Stack is full");
    } else {
      stack[++ top] = value;
    }
  }
  public T peek() {
    if(top == -1) {
      throw new IllegalStateException("Stack is empty");
    }
    else return stack[top];
  }
  public T pop() {
    if(top == -1) {
      throw new IllegalStateException("Stack is empty");
    } 
    else return stack[top--];
  }
}

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🤔

Memory Leak

public T pop() {
  if(top == -1) {
    throw new IllegalStateException("Stack is empty");
  } 
  else return stack[top--];
}

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class Stack<T> {
  private int top;
  int capacity;
  T[] stack;
  public Stack(int inCapacity) {
    capacity = inCapacity;
    stack = (T[]) new Object[capacity];
    top = -1;
  }
  public int size() {
    return top + 1;
  }
  public void push(T value) {
    if(top + 1 == capacity) {
      throw new IllegalStateException("Stack is full");
    } else {
      stack[++ top] = value;
    }
  }
  public T peek() {
    if(top == -1) {
      throw new IllegalStateException("Stack is empty");
    }
    else return stack[top];
  }
  public T pop() {
    if(top == -1) {
      throw new IllegalStateException("Stack is empty");
    } 
    else {
        T result = stack[top--];
        stack[top + 1] = null;
        return result;
    }
  }
}

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When to use Stack?

Invoke a function
Recursion (This is a special case of the first)
DFS (Depth-first Search) (This is a special case of the second)

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Stack in Operating System

Stack and Heap
Stack is Faster
Heap can dynamically allocate space
Objects are all in heap
Functions are in stacks

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static void method(int value) {
    int a = value;
    int[] arr = new int[3];
    method(a + 2);
}

public static void main(String[] args) {
    method(3);
}

🤔

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static void method(int value) {
    if (value > 5) {
        return;
    }
    int a = value;
    int[] arr = new int[3];
    method(a + 2);
}

public static void main(String[] args) {
    method(3);
}

Stack in Operating System

Stack and Heap
Stack is Faster
Heap can dynamically allocate space
Objects are all in heap
Functions are in stacks

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Stack in Operating System

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Queue

LinkedList<String> names = new LinkedList<>();

Java

C++

queue <string> names;

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Queue in Java is an Interface. Usually we use linkedlist to implement features for queues

Basic Operations

size
remove (poll)
add (offer)
element (peek)

size
pop_front
push_back

Java

C++

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Queue

How to use Array to implement a queue?

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What might be the problems

🤔

Array => Queue

class Queue<T> {
    private int front;
    private int rear;
    int capacity;
    int size;
    T[] queue;
    public Queue(int inCapacity) {
        capacity = inCapacity;
        queue = (T[]) new Object[capacity];
        front = 0;
        rear = 0;
        size = 0;
    }
    public int size() {
        return size;
    }
}

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Array => Queue (Cont.d)

class Queue<T> {
    public void add(T value) {
        if (size == capacity) {
            throw new IllegalStateException("Queue is full");

        } else {
            queue[rear] = value;
            rear = (rear + 1) % capacity;
            size++;
        }
    }
    public T remove() {
        T value = null;
        if (size == 0) {
            throw new IllegalStateException("Cannot remove empty queue");
        } else {
            value = queue[front];
            front = (front + 1) % capacity;
            size--;
        }
        return value;
    }
}

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When to use Queue?

BFS (Breadth-first Search)
Priority Queue (Heap)
Multi Task Queue (Design)

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LinkedList Rule Them All

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Head

Tail

pop

(poll)

pollFirst

remove

removeFirst

push

addFirst

offerFirst

peek

(element)

getFirst

peekFirst

pollLast

removeLast

add

(offer)

addAll

addLast

offerLast

peekLast

getLast

java.util.LinkedList

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Practice 1

Implement Queue using Stacks

Use two Stacks to implement a queue

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Implement Queue using Stacks

Use two Stacks to implement a queue

Consider the basic operations of these two data structures

How to minimize the time complexity?

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Implement Queue using Stacks

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Implement Queue using Stacks

class MyQueue {
    LinkedList<Integer> stack = new LinkedList<>();
    // Push element x to the back of queue.
    public void push(int x) {
        LinkedList<Integer> temp = new LinkedList<>();
        while (!stack.isEmpty()) {
            temp.push(stack.pop());
        }
        stack.push(x);
        while (!temp.isEmpty()) {
            stack.push(temp.pop());
        }
    }
    // Removes the element from in front of queue.
    public void pop() {
        stack.pop();
    }
    // Get the front element.
    public int peek() {
        return stack.peek();
    }
    // Return whether the queue is empty.
    public boolean empty() {
        return stack.isEmpty();
    }
}

Big O

O(n)

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Implement Queue using Stacks

class MyQueue {
    Stack<Integer> temp = new Stack<Integer>();
    Stack<Integer> value = new Stack<Integer>();
    // Push element x to the back of queue.
    public void push(int x) {
        value.push(x);
    }
    // Removes the element from in front of queue.
    public void pop() {
        if(temp.isEmpty()) {
            while(!value.isEmpty()) {
                temp.push(value.pop());
            }
        }
        temp.pop();        
    }
    // Get the front element.
    public int peek() {
        if(temp.isEmpty()) {
            while(!value.isEmpty()) {
                temp.push(value.pop());
            }
        }
        return temp.peek();        
    }
    // Return whether the queue is empty.
    public boolean empty() {
        return temp.isEmpty() && value.isEmpty();       
    }
}

Big O

O(1)

Any Problem?

🤔

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Implement Queue using Stacks

class MyQueue {
    Stack<Integer> temp = new Stack<Integer>();
    Stack<Integer> value = new Stack<Integer>();
    // Push element x to the back of queue.
    public void push(int x) {
        value.push(x);
    }
    // Removes the element from in front of queue.
    public void pop() {
        reorder();
        temp.pop();        
    }
    // Get the front element.
    public int peek() {
        reorder();
        return temp.peek();        
    }
    // Return whether the queue is empty.
    public boolean empty() {
        return temp.isEmpty() && value.isEmpty();       
    }
    private void reorder() {
        if(temp.isEmpty()) {
            while(!value.isEmpty()) {
                temp.push(value.pop());
            }
        }
    }
}

Practice 2

a stack with max()

Instead normal stack operation, this stack will have a max() function that could return the current max value of the stack

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a stack with max()

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One stack cannot tell the max in O(1)

Need another stack to store the max

time <=> space

class MaxStack {
  Stack element;
  Stack maxValue;
  public MaxStack(int inCapacity) {
    element = new Stack(inCapacity);
    maxValue = new Stack(inCapacity);
  }
  public int size() {
    return element.size();
  }
  public void push(int value) {
    int currentMax;
    if(element.size() == 0) {
      currentMax = Integer.MIN_VALUE;
    } else {
      currentMax = maxValue.peek();
    }
    element.push(value);
    if(currentMax > value) {
      maxValue.push(currentMax);
    }
    else maxValue.push(value);
  }
  public int pop() {
    maxValue.pop();
    return element.pop();
  }
  public int getMax() {
    return maxValue.peek();
  }
}

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EmptyStackException

store < n extra

🤔

class MaxStack {
  Stack element;
  Stack maxValue;
  public MaxStack(int inCapacity) {
    element = new Stack(inCapacity);
    maxValue = new Stack(inCapacity);
  }
  public int size() {
    return element.size();
  }
  public void push(int value) {
    element.push(value);
    if(maxValue.isEmpty() || maxValue.peek() <= value) {
        maxValue.push(value);
    }
  }
  public int pop() {
    if (!maxValue.isEmpty() && maxValue.peek() == element.peek()) {
        maxValue.pop();
    }
    return element.pop();
  }
  public int getMax() {
    return maxValue.peek();
  }
}

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🤔

only use one stack

~~store < n extra~~

!maxValue.isEmpty()

class MaxStack {
    private class Element {
        int val;
        int max;
        public Element(int val, int max) {
            this.val = val;
            this.max = max;
        }
    }
    Stack<Element> stack = new Stack<>();
    public int size() {
        return stack.size();
    }
    public void push(int x) {
        int max = (stack.isEmpty()) ? x : Math.max(x, stack.peek().max);
        stack.push(new Element(x, max));
    }
    public int pop() {
        return stack.pop().val;
    }
    public int getMax() {
        return stack.peek().max;
    }
}

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Practice 3

Valid Parentheses

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Valid Parentheses

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We need to use the stack to store the information: Which left parentheses has not been matched yet.

Stack can help store every unused parentheses, no matter when the parentheses appears

public boolean isValid(String s) {
	Stack<Character> pair = new Stack<Character>();
	int n = s.length();
	if(n == 0) return true;
	for(int i = 0; i < n; i ++) {
		if(s.charAt(i) == '(' || s.charAt(i) == '[' || s.charAt(i) == '{')  {
			pair.push(s.charAt(i));
		}
		else if(s.charAt(i) == ')') {
			if(pair.isEmpty()) return false;
			if(pair.pop() != '(') return false;
		}
		else if(s.charAt(i) == ']') {
			if(pair.isEmpty()) return false;
			if(pair.pop() != '[') return false;
		}
		else if(s.charAt(i) == '}') {
			if(pair.isEmpty()) return false;
			if(pair.pop() != '{') return false;
		}
	}
	if(pair.isEmpty()) return true;
	return false;
}

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🤔

public boolean isValid(String s) {
	Stack<Character> pair = new Stack<Character>();
	int n = s.length();
	if(n == 0) return true;
	for(int i = 0; i < n; i ++) {
		if(s.charAt(i) == '(' || s.charAt(i) == '[' || s.charAt(i) == '{')  {
			pair.push(s.charAt(i));
		}
		else if(s.charAt(i) == ')') {
			if(pair.isEmpty() || pair.pop() != '(') return false;
		}
		else if(s.charAt(i) == ']') {
			if(pair.isEmpty() || pair.pop() != '[') return false;
		}
		else if(s.charAt(i) == '}') {
			if(pair.isEmpty() || pair.pop() != '{') return false;
		}
	}
	return pair.isEmpty();
}

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🤔

messy?

public boolean isValid(String s) {
    LinkedList<Character> stack = new LinkedList<>();
    Map<Character, Character> hm = new HashMap<>();
    hm.put(')', '(');
    hm.put(']', '[');
    hm.put('}', '{');
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (hm.containsKey(c)) {
            if (stack.isEmpty() || !hm.get(c).equals(stack.pop())) {
                return false;
            }
        } else {
            stack.push(c);
        }
    }
    return stack.isEmpty();
}

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Practice 4

Bad Hair Day

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=       
=       =        Cows facing right -->
=     = =
=   = = =
= = = = = =
1 2 3 4 5 6

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All the cows are facing right, and they can only see other cows in front of it and shorter than itself. It can be blocked by other taller cows.

Example: 6 2 3 1 7 4 Output: 5

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SUM(each cow can see how many other cows?) = SUM(each cow can be seen by how many other cows)

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SUM(each cow can see how many other cows?) = SUM(each cow can be seen by how many other cows)

How to use Stack?

How could Stack help record this?

Only allow cows whose height is lower than you into the stack

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How to use Stack?

Example: 6 2 3 1 7 4 Result: 5

Stack: 6

Stack: 6 2

Stack: 6 3

Stack: 6 3 1

Stack: 7

Stack: 7 4

Stack: 7

Stack: empty

Add 6

Add 2

Pop 2 Add 3

Add 1

Pop 6 3 1 Add 7

Add 4

Pop 4

Pop 7

  public static int seeCows(int[] cow) {
    int length = cow.length;
    int totalNum = 0;
    Stack<Integer> cowStack = new Stack<Integer>();
    for(int i = 0; i < length; i ++){
      if(cowStack.size() == 0 || cowStack.peek() > cow[i]) {
        cowStack.push(cow[i]);
      }
      else {
        while(cowStack.size() > 0 && cowStack.peek() <= cow[i]) {
          cowStack.pop();
          totalNum += cowStack.size();
        }
        cowStack.push(cow[i]);
      }
    }
    while(cowStack.size() > 0) {
      cowStack.pop();
      totalNum += cowStack.size();
    }
    return totalNum;
  }

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Big O?

O(n)

Monotonic Stack

单调栈

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Find an element's left/right larger/smaller element in O(n) time.

Practice 5

Largest Rectangle in Histogram

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Largest Rectangle in Histogram

Similar thought: How to use stack to find the right border?

Example: 2 1 5 6 2 3

Only put rectangle higher than peek into stack

The element before it is the left border

The element pushes it out is the right border

Monotonically increasing or decreasing?

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Largest Rectangle in Histogram

Example: 2 1 5 6 2 3

Stack: 2

Stack: 1

Stack: 1 5

Stack: 1 5 6

Stack: 1 2

Stack: 1 2 3

Stack: 1 2

Stack: 1

Stack: Empty

Add 2

Pop 2 Add 1

Add 5

Add 6

Pop 6 5 Add 2

Add 3

Pop 3

Pop 2

Pop 1

2*1

6*1, 5*2

3*1

2*4

1*6

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Largest Rectangle in Histogram

Example: 2 1 5 6 2 3

How do we know the width?

Instead of directly putting height into stack, we put the index. Anyway height can be accessed by the index easily. And using index, we can know the width

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Largest Rectangle in Histogram

0 1 2 3 4 5

2 1 5 6 2 3

Stack: 0

Stack: 1

Stack: 1 2

Stack: 1 2 3

Stack: 1 4

Stack: 1 4 5

Stack: 1 4

Stack: 1

Stack: empty

Add 0

Pop 0 Add 1

Add 2

Add 3

Pop 2 3 Add 4

Add 5

Pop 5

Pop 4

Pop 1

2*1

6*1, 5*2

3*1

2*4

1*6

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public int largestRectangleArea(int[] height) {
    int area = 0;
    jStack<Integer> stack = new Stack<Integer>();
    for (int i = 0; i < height.length; i++) {
        if (stack.empty() || height[stack.peek()] < height[i]) {
            stack.push(i);
        } else {
            int start = stack.pop();
            int width = stack.empty() ? i : i - stack.peek() - 1;
            area = Math.max(area, height[start] * width);
            i--;
        }
    }
    while (!stack.empty()) {
        int start = stack.pop();
        int width = stack.empty() ? height.length : height.length - stack.peek() - 1;
        area = Math.max(area, height[start] * width);
    }
    return area;
}

🤔

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public int largestRectangleArea(int[] height) {
    int n = height.length;
    int result = 0;
    Stack<Integer> stack = new Stack<>();
    for (int i = 0; i <= n; i++) {
        int cur = (i == n) ? 0 : height[i];
        while (!stack.isEmpty() && cur < height[stack.peek()]) {
            int area = height[stack.pop()] * (stack.isEmpty() ? i : i - 1 - stack.peek());
            result = Math.max(area, result);
        }
        stack.push(i);
    }
    return result;
}

cur <= height[stack.peek()]

Exist Balance Number

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Given an array of integers.

Is there a number k in this array that

k is larger than all the numbers before it and smaller than all the numbers after it?

[1,2,3,4,5] -> true

[6,2,3,8,10,9] -> true

[4,3,5,2] -> false

[2] -> true

extra practice

Return all balance numbers

Homework 6.1

Simplify Path

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Homework 6.2

Implementing Stack using Queues

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Homework 6.3

Trapping Rain Water

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Explain

Homework 6 (optional)

Container With Most Water

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Explain

Homework 6 (more)

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Basic Calculator
Evaluate Reverse Polish Notation
Flatten Nested List Iterator
132 Pattern (optional)
Q -> S; S -> Q; Array -> Q, S
Implement Java LL with Node
- Front & Rear
  - popFront, popRear
  - pushFront, pushRear
  - peekFront, peekRear
  - isEmpty, size