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 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.
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:
- N is in range [1,200].
- M[i][i] = 1 for all students.
- If M[i][j] = 1, then M[j][i] = 1.
一個班級有N個學生，他們中的一些相互之間是朋友，朋友關系可以傳遞。如果A和B是直接好友，B和C是直接好友，那么即使A和C是間接好友，他們三人屬於一個朋友圈。朋友圈是由直接好友和間接好友構成。給一個N * N的矩陣M表示學生之間的朋友關系。M[i][j] = 1表示學生i和學生j是直接朋友，否則不是。輸出學生中的朋友圈個數。
思路：遍歷矩陣里的點，如果是1，就找出和他是朋友的點和朋友的朋友的點，標記好。最后找出朋友圈的個數。
解法1: DFS
解法2: BFS
解法3: Union Find
解法4: 並查集（Disjoint-set data structure） 把有有關系的人放到一個集合里，然后計算集合的個數。
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++) {
// 未被訪問過且是鄰接點,注意是node的鄰接點
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:
vset.add(x)
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:
vset.add(x)
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:
group.add(j)
allNodes = set()
for i in xrange(len(M)):
allNodes.add(i)
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):
visited.add(root)
allNodes.discard(root)
unvisited = set()
for node in hm.get(root):
if node not in visited:
unvisited.add(node)
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: 並查集
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++: 並查集
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;
}
類似題目：
[LeetCode] 323. Number of Connected Components in an Undirected Graph 無向圖中的連通區域的個數
[LeetCode] 200. Number of Islands 島嶼的數量
[LeetCode] 305. Number of Islands II 島嶼的數量 II
All LeetCode Questions List 題目匯總