Initialization
1
2
3
4
5
6
7
1
2
3
4
5
6
7
union(2,4), union(5,7)
union(4,7)
1
2
3
4
5
6
7
find(7) = 2
class UnionFind {
List<Integer> list;
public UnionFind(int n) {
list = new ArrayList<>();
for (int i = 0; i < n; i ++) {
list.add(i);
}
}
boolean union(int a, int b) {
int ancesterA = find(a), ancesterB = find(b);
if (ancesterA == ancesterB) return false; // need not to union.
else {
list.set(ancesterB, ancesterA);
return true;
}
}
int find(int k) {
int i = k;
while (i != list.get(i)) {
i = list.get(i); //Here i is the root.
}
return i;
}
}class UnionFind:
def __init__(self, n):
self.list = [i for i in range(n)]
def union(self, a, b):
ancestor_a = self.find(a)
ancestor_b = self.find(b)
if ancestor_a == ancestor_b:
return False
else:
self.list[ancestor_b] = ancestor_a
return True
def find(self, k):
i = k
while i != self.list[i]:
i = self.list[i] # Here i is the root.
return iGiven a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.
A region is captured by flipping all 'O's into 'X's in that surrounded region.
X X X X
X O O X
X X O X
X O X X
X X X X
X X X X
X X X X
X O X X
Before we use BFS or DFS to solve this question
But Union Find can also solve this question
class Solution {
List<Integer> list = new ArrayList<>();
List<Boolean> edge = new ArrayList<>();
public void solve(char[][] board) {
if (board.length == 0 || board[0].length == 0) return;
int m = board.length;
int n = board[0].length;
for (int i = 0; i < m * n; i ++) {
int x = i / n;
int y = i % n;
if ((x == 0 || x == m - 1 || y == 0 || y == n - 1) && (board[x][y] == 'O')) {
edge.add(true);
} else {
edge.add(false);
}
list.add(i);
}
for (int i = 0; i < m * n; i ++) {
int x = i / n;
int y = i % n;
if (x < m - 1 && board[x][y] == board[x + 1][y]) {
union(i, i + n);
}
if (y < n - 1 && board[x][y] == board[x][y+1]) {
union(i, i + 1);
}
}
for (int i = 0; i < m * n; i ++) {
int x = i / n;
int y = i % n;
if (board[x][y] == 'O' && !edge.get(find(i))) {
board[x][y] = 'X';
}
}
} boolean union(int a, int b) {
int ancesterA = find(a), ancesterB = find(b);
if (ancesterA == ancesterB) return false; // need not to union.
else {
list.set(ancesterB, ancesterA);
if (edge.get(ancesterB)) {
edge.set(ancesterA, true);
}
return true;
}
}
int find(int k) {
int i = k;
while (i != list.get(i)) {
i = list.get(i); //Here i is the root.
}
return i;
}
}class Solution:
def __init__(self):
self.list = []
self.edge = []
def solve(self, board: List[List[str]]) -> None:
if len(board) == 0 or len(board[0]) == 0:
return
m = len(board)
n = len(board[0])
for i in range(m * n):
x = i // n
y = i % n
if (x == 0 or x == m - 1 or y == 0 or y == n - 1) and (board[x][y] == 'O'):
self.edge.append(True)
else:
self.edge.append(False)
self.list.append(i)
for i in range(m * n):
x = i // n
y = i % n
if x < m - 1 and board[x][y] == board[x + 1][y]:
self.union(i, i + n)
if y < n - 1 and board[x][y] == board[x][y + 1]:
self.union(i, i + 1)
for i in range(m * n):
x = i // n
y = i % n
if board[x][y] == 'O' and not self.edge[self.find(i)]:
board[x][y] = 'X' def union(self, a, b):
ancestor_a = self.find(a)
ancestor_b = self.find(b)
if ancestor_a == ancestor_b:
return False
else:
self.list[ancestor_b] = ancestor_a
if self.edge[ancestor_b]:
self.edge[ancestor_a] = True
return True
def find(self, k):
i = k
while i != self.list[i]:
i = self.list[i] # Here i is the root.
return iThere are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a directfriend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.
Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are directfriends with each other, otherwise not. And you have to output the total number of friend circles among all the students.
Input:
[[1,1,0],
[1,1,0],
[0,0,1]]
Output: 2
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle.
The 2nd student himself is in a friend circle. So return 2.
It is very intuitive that Union Find can solve the problem. We just need to keep doing the union and count after all how many sets there are.
class Solution {
List<Integer> list = new ArrayList<>();
void union(int a, int b) {
int ancesterA = find(a), ancesterB = find(b);
if (ancesterA == ancesterB) return; // need not to union.
else {
list.set(ancesterB, ancesterA);
}
}
int find(int k) {
int i = k;
while (i != list.get(i)) {
i = list.get(i); //Here i is the root.
}
return i;
}
public int findCircleNum(int[][] M) {
int n = M.length;
int count = n;
for (int i = 0; i < n; i ++) {
list.add(i);
}
for (int i = 0; i < n; i ++) {
for (int j = 0; j < i; j ++) {
if (M[i][j] == 1 && find(i) != find(j)) {
union(i, j);
count --;
}
}
}
return count;
}
}class Solution:
def __init__(self):
self.list = []
def union(self, a, b):
ancestor_a = self.find(a)
ancestor_b = self.find(b)
if ancestor_a == ancestor_b:
return
else:
self.list[ancestor_b] = ancestor_a
def find(self, k):
i = k
while i != self.list[i]:
i = self.list[i] # Here i is the root.
return i
def findCircleNum(self, M):
n = len(M)
count = n
for i in range(n):
self.list.append(i)
for i in range(n):
for j in range(i):
if M[i][j] == 1 and self.find(i) != self.find(j):
self.union(i, j)
count -= 1
return count