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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

  • 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); } } } } 
  • 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; } }; 
  • 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 
  • 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 { dfs(j) } } } for i, v := range vis { if !v { ans++ dfs(i) } } return } 
  • function findCircleNum(isConnected: number[][]): number { const n = isConnected.length; const vis: boolean[] = new Array(n).fill(false); const dfs = (i: number) => { vis[i] = true; for (let j = 0; j < n; ++j) { if (!vis[j] && isConnected[i][j]) { dfs(j); } } }; let ans = 0; for (let i = 0; i < n; ++i) { if (!vis[i]) { dfs(i); ++ans; } } return ans; } 
  • impl Solution { fn dfs(is_connected: &mut Vec<Vec<i32>>, vis: &mut Vec<bool>, i: usize) { vis[i] = true; for j in 0..is_connected.len() { if vis[j] || is_connected[i][j] == 0 { continue; } Self::dfs(is_connected, vis, j); } } pub fn find_circle_num(mut is_connected: Vec<Vec<i32>>) -> i32 { let n = is_connected.len(); let mut vis = vec![false; n]; let mut res = 0; for i in 0..n { if vis[i] { continue; } res += 1; Self::dfs(&mut is_connected, &mut vis, i); } res } }

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