547. Number of Provinces

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]

 A:

一开始理解错误。 以为是关于edge这个matrix的 岛屿的数量。

但是，这里，从a->b 需要再找每个从b开始的边的dfs

class Solution {

public:

int findCircleNum(vector<vector<int>>& isConnected) {

int n = isConnected.size();

int res = 0;

for (int i = 0; i < n; i++) {

for (int j = 0; j < n; j++) {

if (isConnected[i][j]) {

dfs(isConnected, i, j);

res++;

}

}

}

return res;

}

private:

void dfs(vector<vector<int>>& isConnected, int i, int j) {

int n = isConnected.size();

isConnected[i][j] = 0;

for (int k = 0; k < n; k++) {

if (isConnected[j][k] == 1)

dfs(isConnected, j, k);

}

}

};