题目：岛屿的个数

给一个01矩阵，求不同的岛屿的个数。

0代表海，1代表岛，如果两个1相邻，那么这两个1属于同一个岛。我们只考虑上下左右为相邻。

您在真实的面试中是否遇到过这个题？

Yes

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样例

在矩阵：

[  [1, 1, 0, 0, 0],  [0, 1, 0, 0, 1],  [0, 0, 0, 1, 1],  [0, 0, 0, 0, 0],  [0, 0, 0, 0, 1]]

中有 3 个岛.

相关题目 Expand

3(union-find) 困难 岛屿的个数II 11 % 

public class Solution {
    /**
     * @param grid a boolean 2D matrix
     * @return an integer
     */
    public int numIslands(boolean[][] grid) {
        // Write your code here
                  if (grid == null || grid.length == 0) {
               return 0;
          }
          int m = grid.length;
          int n = grid[0].length;
          boolean[][] visited = new boolean[m][n];// true表示访问过，false表示没有访过
          for(int i=0;i<m;i++){
               for(int j=0;j<n;j++){
                    visited[i][j] = false;
               }
          }
          int res = 0;
          for (int i = 0; i < m; i++) {
               for (int j = 0; j < n; j++) {
                    if (grid[i][j] && !visited[i][j]) {
                         DPSlands(grid, visited, i, j);
                         res++;
                    }
               }
          }
          return res;
     }

     public void DPSlands(boolean[][] grid, boolean[][] visited, int i, int j) {
          if (i < 0 || i >= grid.length) {
               return;
          }
          if (j < 0 || j >= grid[0].length) {
               return;
          }
          if (!grid[i][j] || visited[i][j]) {
               return;
          }
          visited[i][j] = true;
          DPSlands(grid, visited, i - 1, j);
          DPSlands(grid, visited, i, j - 1);
          DPSlands(grid, visited, i + 1, j);
          DPSlands(grid, visited, i, j + 1);
    }
}