433. Number of Islands【LintCode java】

Description

Given a boolean 2D matrix, 0 is represented as the sea, 1 is represented as the island. If two 1 is adjacent, we consider them in the same island. We only consider up/down/left/right adjacent.

Find the number of islands.

Example

Given graph:

[
  [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]
]

return 3.

解题：算岛屿的个数。因为连接在一起的陆地都算是岛屿，所以当确定有一一个格子是陆地时，它周围的格子即使是陆地，也不能重复计算，而应该当成海洋（对应于下面代码中的dfs函数）。代码如下：

public class Solution {
    /**
     * @param grid: a boolean 2D matrix
     * @return: an integer
     */
    public int numIslands(boolean[][] grid) {
        // write your code here
        int res = 0;
        int m = grid.length;
        if(m == 0)
            return 0;
        int n = grid[0].length;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == true){
                    res++;
                    dfs(grid, i, j);
                }
            }
        }
        return res;
    }
    public void dfs(boolean[][]grid, int i, int j){
        if(i < 0 || j < 0 || i >= grid.length || j >= grid[0].length){
            return;
        }
        if(grid[i][j] == true){
            grid[i][j] = false;
            dfs(grid, i-1, j);
            dfs(grid, i+1, j);
            dfs(grid, i, j-1);
            dfs(grid, i, j+1);
        }
    }
}