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
和数字岛屿的区别：
这个例子
0,0 0,3
1,1 1,2
2,1 2,2 2,3
3,0 3,2 3,3
0123全都是相通的，所以pyq数量=1。但是岛屿数量是4.
您有一个 NxN 矩阵，但总共只有 N 个朋友。所以一行只loop一次就行了。
别的特点：本题对角元素全为1。
教训：题目改名字了，最好重新搜索一下。一定要小心再小心！
做法是修改一下dfs
public class Solution {
private int n;
private int m;
public int numIslands(char[][] grid) {
int count = 0;
n = grid.length;
if (n == 0) return 0;
m = grid[0].length;
for (int i = 0; i < n; i++){
for (int j = 0; j < m; j++)
if (grid[i][j] == '1') {
DFSMarking(grid, i, j);
++count;
}
}
return count;
}
private void DFSMarking(char[][] grid, int i, int j) {
if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] != '1') return;
grid[i][j] = '0';
for (int d = 0; d < n; d++)
dfs(grid, d, j);
for (int d = 0; d < m; d++)
dfs(grid, i, d);
}
}