200. Number of Islands (Graph)

DFS把能遍历到的是1的都变成0，这样就能识别出一个岛。

 

 1 class Solution {
 2     public int numIslands(char[][] grid) {
 3         if(grid == null) return 0;
 4         if(grid.length == 0) return 0;
 5         int n = grid.length;
 6         int m = grid[0].length;
 7         int count = 0;
 8         for(int i = 0; i < n; i++) {
 9             for(int j = 0; j < m; j++) {
10                 if(grid[i][j] == '1') {
11                     dfs(grid, i, j);
12                     count++;
13                     
14                 }
15                 
16             }
17         }
18         return count;
19         
20         
21     }
22     
23     public void dfs(char[][] grid, int i, int j) {
24         if(i >= 0 && j >= 0 && i < grid.length && j < grid[0].length && grid[i][j] == '1') { //grid[i][j]判断条件要写在最后 不然会outofbound
25             grid[i][j] = '0';
26             dfs(grid, i - 1, j);
27             dfs(grid, i + 1, j);
28             dfs(grid, i, j - 1);
29             dfs(grid, i, j + 1);
30         }
31         
32     }
33 }

 

posted @ 2018-08-17 03:01  jasoncool1  阅读(177)  评论(0)    收藏  举报