# [LeetCode] 547. Friend Circles 朋友圈

2021年09月15日 阅读数：2

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.html

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

Example 1:node

```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:python

```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:app

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.

Java: DFS

```public class Solution {
public int findCircleNum(int[][] M) {
int res = 0;
int[] visited = new int[M.length];
for (int i = 0; i < M.length; i++) {
if (visited[i] == 0) {
res++;
dfs(M, visited, i);
}
}
return res;
}

private void dfs(int[][] M, int[] visited, int i) {
visited[i] = 1;
for (int j = 0; j < M.length; j++) {
if (M[i][j] == 1 && visited[j] == 0) {
dfs(M, visited, j);
}

}
}
}
```

Java: BFS

```Queue<Integer> q = new LinkedList<>();
public void bfs(int[][] M, int[] visited, int i) {
// visit[i];
q.offer(i);
visited[i] = 1;
while (!q.isEmpty()) {
int node = q.poll();
for (int j = 0; j < M.length; j++) {
// 未被访问过且是邻接点,注意是node的邻接点
if (visited[j] == 0 && M[node][j] == 1) {
// visit[j];
q.offer(j);
visited[j] = 1;
}
}
}

}
public int findCircleNum(int[][] M) {
int[] visited = new int[M.length];
int count = 0;
for (int i = 0; i < M.length; i++) {
if (visited[i] == 0) {
bfs(M, visited, i);
count++;
}
}
return count;
} 　　```

Python: DFS

```class Solution(object):
def findCircleNum(self, M):
"""
:type M: List[List[int]]
:rtype: int
"""
cnt, N = 0, len(M)
vset = set()
def dfs(n):
for x in range(N):
if M[n][x] and x not in vset:
dfs(x)
for x in range(N):
if x not in vset:
cnt += 1
dfs(x)
return cnt
```

Python: BFS

```class Solution(object):
def findCircleNum(self, M):
"""
:type M: List[List[int]]
:rtype: int
"""
cnt, N = 0, len(M)
vset = set()
def bfs(n):
q = [n]
while q:
n = q.pop(0)
for x in range(N):
if M[n][x] and x not in vset:
q.append(x)
for x in range(N):
if x not in vset:
cnt += 1
bfs(x)
return cnt　```

Python: DFS

```class Solution(object):
def findCircleNum(self, M):
"""
:type M: List[List[int]]
:rtype: int
"""
res = 0
hm = dict()
for i in xrange(len(M)):
for j in xrange(len(M)):
group = hm.setdefault(i, set())
if M[i][j] == 1:

allNodes = set()
for i in xrange(len(M)):
while len(allNodes) != 0:
res += 1
root = None
for node in allNodes:
root = node
break
self.dfs(root, set(), allNodes, hm)

return res

def dfs(self, root, visited, allNodes, hm):
unvisited = set()
for node in hm.get(root):
if node not in visited:
for node in unvisited:
self.dfs(node, visited, allNodes, hm)　　```

Python: Union Find

```class Solution(object):
def findCircleNum(self, M):
"""
:type M: List[List[int]]
:rtype: int
"""

class UnionFind(object):
def __init__(self, n):
self.set = range(n)
self.count = n

def find_set(self, x):
if self.set[x] != x:
self.set[x] = self.find_set(self.set[x])  # path compression.
return self.set[x]

def union_set(self, x, y):
x_root, y_root = map(self.find_set, (x, y))
if x_root != y_root:
self.set[min(x_root, y_root)] = max(x_root, y_root)
self.count -= 1

circles = UnionFind(len(M))
for i in xrange(len(M)):
for j in xrange(len(M)):
if M[i][j] and i != j:
circles.union_set(i, j)
return circles.count
```

Python: 并查集

```class Solution(object):
def findCircleNum(self, M):
"""
:type M: List[List[int]]
:rtype: int
"""
N = len(M)
f = range(N)

def find(x):
while f[x] != x: x = f[x]
return x

for x in range(N):
for y in range(x + 1, N):
if M[x][y]: f[find(x)] = find(y)
return sum(f[x] == x for x in range(N))　```

C++: DFS

```class Solution {
public:
void dfs(vector<int> & v, vector<vector<int>>& M, int line) {
for (int i = 0; i < M[line].size(); i++) {
if (M[line][i] == 1 && v[i] == 0) {
v[i] = 1;
dfs(v, M, i);
}
}
}
int findCircleNum(vector<vector<int>>& M) {
vector<int> v(M.size(), 0);
int count = 0;
for (int i = 0; i < M.size(); i++) {
if (v[i] == 0) {
dfs(v, M, i);
count++;
}
}
return count;
}
};
```

C++: BFS

```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;
}
};　```

C++: Union Find

```class Solution {
public:
int findCircleNum(vector<vector<int>>& M) {
UnionFind circles(M.size());
for (int i = 0; i < M.size(); ++i) {
for (int j = 0; j < M[i].size(); ++j) {
if (M[i][j] && i != j) {
circles.union_set(i, j);
}
}
}
return circles.size();
}

private:
class UnionFind {
public:
UnionFind(const int n) : set_(n), count_(n) {
iota(set_.begin(), set_.end(), 0);
}

int find_set(const int x) {
if (set_[x] != x) {
set_[x] = find_set(set_[x]);  // Path compression.
}
return set_[x];
}

void union_set(const int x, const int y) {
int x_root = find_set(x), y_root = find_set(y);
if (x_root != y_root) {
set_[min(x_root, y_root)] = max(x_root, y_root);
--count_;
}
}

int size() const {
return count_;
}

private:
vector<int> set_;
int count_;
};
};
```

C++: 并查集

```struct DisjointSet{
int par;
int rank;
};
#define maxn 222
struct DisjointSet ds[maxn];

void init()
{
for(int i = 0;i < maxn;i++){
ds[i].par = i;
ds[i].rank = 1;
}
}

int find(int x)
{
if(x == ds[x].par) return  x;
return ds[x].par = find(ds[x].par);
}

void _union(int x, int y)
{
x = find(x);
y = find(y);
if(x == y) return ;
if(ds[x].rank < ds[y].rank)
ds[x].par = y;
else{
if(ds[x].rank == ds[y].rank)
ds[x].rank++;
ds[y].par = x;
}
}

int same(int x, int y)
{
return find(x) == find(y);
}

int findCircleNum(int** M, int MRowSize, int MColSize) {
init();
for(int i = 0;i < MRowSize;i++){
for(int j = i + 1;j < MColSize;j++){
if(M[i][j] && M[j][i]) _union(i,j);
}
}
int count = 0;
for(int i = 0;i < MRowSize;i++){
if(ds[i].par == i) count++;
}
return count;
}
```

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