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547. Number of Provinces

Description

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.

A province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.

Return the total number of provinces.

 

Example 1:

Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2

Example 2:

Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3

 

Constraints:

  • 1 <= n <= 200
  • n == isConnected.length
  • n == isConnected[i].length
  • isConnected[i][j] is 1 or 0.
  • isConnected[i][i] == 1
  • isConnected[i][j] == isConnected[j][i]

Solutions

Solution 1

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class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        def dfs(i: int):
            vis[i] = True
            for j, x in enumerate(isConnected[i]):
                if not vis[j] and x:
                    dfs(j)

        n = len(isConnected)
        vis = [False] * n
        ans = 0
        for i in range(n):
            if not vis[i]:
                dfs(i)
                ans += 1
        return ans

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class Solution {
    private int[][] g;
    private boolean[] vis;

    public int findCircleNum(int[][] isConnected) {
        g = isConnected;
        int n = g.length;
        vis = new boolean[n];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {
                dfs(i);
                ++ans;
            }
        }
        return ans;
    }

    private void dfs(int i) {
        vis[i] = true;
        for (int j = 0; j < g.length; ++j) {
            if (!vis[j] && g[i][j] == 1) {
                dfs(j);
            }
        }
    }
}

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class Solution {
public:
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        int ans = 0;
        bool vis[n];
        memset(vis, false, sizeof(vis));
        function<void(int)> dfs = [&](int i) {
            vis[i] = true;
            for (int j = 0; j < n; ++j) {
                if (!vis[j] && isConnected[i][j]) {
                    dfs(j);
                }
            }
        };
        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {
                dfs(i);
                ++ans;
            }
        }
        return ans;
    }
};

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func findCircleNum(isConnected [][]int) (ans int) {
    n := len(isConnected)
    vis := make([]bool, n)
    var dfs func(int)
    dfs = func(i int) {
        vis[i] = true
        for j, x := range isConnected[i] {
            if !vis[j] && x == 1 {