该题目要算出矩阵中有多少个数值为“1”的数据块。主要思路是对矩阵进行遍历，遇到“1”的元素的时候，对元素进行深度遍历，并将“1”元素改为非“1”。
代码如下：
1 class Solution {
2 public int numIslands(char[][] grid) {
3 int res = 0;
4 if(grid == null || grid.length == 0){
5 return res;
6 }
7
8 int m = grid.length, n = grid[0].length;
9
10 for( int i = 0 ; i < m ; i++ ){
11 for( int j = 0 ; j < n ; j++){
12 if(grid[i][j] == '1'){
13 res ++;
14 occupyIsland(grid, i, j);
15 }
16 }
17 }
18
19 return res;
20 }
21
22 private void occupyIsland(char[][] grid, int i, int j){
23 grid[i][j] = '2';
24
25 int m = grid.length, n = grid[0].length;
26 if( i+1 < m && grid[i+1][j] == '1'){
27 occupyIsland(grid, i+1, j);
28 }
29 if( j+1 < n && grid[i][j+1] == '1'){
30 occupyIsland(grid, i, j+1);
31 }
32 if( i-1 >= 0 && grid[i-1][j] == '1'){
33 occupyIsland(grid, i-1, j);
34 }
35 if( j-1 >= 0 && grid[i][j-1] == '1'){
36 occupyIsland(grid, i, j-1);
37 }
38 }
39 }
END