[LeetCode 200] Number of Islands

标签：java   算法   dfs   

题目链接：number-of-islands

/**
 * 
		Given a 2d grid map of '1's (land) and '0's (water), 
		count the number of islands. An island is surrounded by water 
		and is formed by connecting adjacent lands horizontally or vertically. 
		You may assume all four edges of the grid are all surrounded by water.
		
		Example 1:
		
		11110
		11010
		11000
		00000
		Answer: 1
		
		Example 2:
		
		11000
		11000
		00100
		00011
		Answer: 3
 *
 */

public class NumberOfIslands {

//	45 / 45 test cases passed.
//	Status: Accepted
//	Runtime: 298 ms
//	Submitted: 0 minutes ago

	////时间复杂度O(n ^ 2)，空间复杂度O(1)
    public int numIslands(char[][] grid) {
    	int m = grid.length;
        if(m == 0) return 0;
        int n = grid[0].length;
        int islands = 0;
        for(int i = 0; i < m; i ++) {
        	for (int j = 0; j < n; j++) {
				if(grid[i][j] == '1') {
					islands ++;
					dfs(grid, i, j);
				}
			}
        }
    	return islands;
    }
    public void dfs(char[][] grid, int i, int j) {
    	if(i < 0 || i == grid.length || j < 0 || j == grid[0].length)
    		return;					//验证在矩阵范围内
    	
    	if(grid[i][j] == '1') {
    		grid[i][j] = '0';		//标记为已访问
    		dfs(grid, i - 1, j);	//左
    		dfs(grid, i + 1, j);	//右
    		dfs(grid, i, j - 1);	//上
    		dfs(grid, i, j + 1);	//下
    	}
    }
	public static void main(String[] args) {
		// TODO Auto-generated method stub

	}

}

[LeetCode 200] Number of Islands

标签：java   算法   dfs   

原文地址：http://blog.csdn.net/ever223/article/details/45062259