# 547. Friend Circles

There 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 direct friend 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 ithand jth students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.

Example 1:

``````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.
``````

Example 2:

``````Input:
[[1,1,0],
[1,1,1],
[0,1,1]]
Output: 1
Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends,
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
``````

Note:

1. N is in range [1,200].
2. M[i][i] = 1 for all students.
3. If M[i][j] = 1, then M[j][i] = 1.

``````class Solution {
public:
int findCircleNum(vector<vector<int>>& M) {
int n = M.size(), res = 0;
vector<bool> visited(n, false);
for (int i = 0; i < n; ++i) {
if (visited[i]) continue;
helper(M, i, visited);
++res;
}
return res;
}
void helper(vector<vector<int>>& M, int k, vector<bool>& visited) {
visited[k] = true;
for (int i = 0; i < M.size(); ++i) {
if (!M[k][i] || visited[i]) continue;
helper(M, i, visited);
}
}
};
``````

``````class Solution {
public:
int findCircleNum(vector<vector<int>>& M) {
int n = M.size(), res = 0;
vector<bool> visited(n, false);
queue<int> q;
for (int i = 0; i < n; ++i) {
if (visited[i]) continue;
q.push(i);
while (!q.empty()) {
int t = q.front(); q.pop();
visited[t] = true;
for (int j = 0; j < n; ++j) {
if (!M[t][j] || visited[j]) continue;
q.push(j);
}
}
++res;
}
return res;
}
};
``````

``````class Solution {
public:
int findCircleNum(vector<vector<int>>& M) {
int n = M.size(), res = n;
vector<int> root(n);
for (int i = 0; i < n; ++i) root[i] = i;
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (M[i][j] == 1) {
int p1 = getRoot(root, i);
int p2 = getRoot(root, j);
if (p1 != p2) {
--res;
root[p2] = p1;
}
}
}
}
return res;
}
int getRoot(vector<int>& root, int i) {
while (i != root[i]) {
root[i] = root[root[i]];
i = root[i];
}
return i;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/547

Accounts Merge

Redundant Connection II

Redundant Connection

Number of Islands II

Graph Valid Tree

Number of Connected Components in an Undirected Graph

Similar String Groups

https://leetcode.com/problems/friend-circles/

https://leetcode.com/problems/friend-circles/discuss/101440/c-bfs

https://leetcode.com/problems/friend-circles/discuss/101338/Neat-DFS-java-solution

https://leetcode.com/problems/friend-circles/discuss/101387/Easy-Java-Union-Find-Solution

LeetCode All in One 题目讲解汇总(持续更新中…)

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