leetcode-200 Number of Islands

最新推荐文章于 2018-02-04 21:39:37 发布

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本文介绍了一种使用深度优先遍历来计算二维网格中岛屿数量的方法，通过修改网格值并避免重复访问，有效地解决了岛屿问题。

问题描述：

Given a 2dgrid map of'1's (land) and'0's (water),count the number of islands. An island is surrounded by water and is formed byconnecting adjacent lands horizontally or vertically. You may assume all fouredges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

Credits:
Special thanks to @mithmatt for adding this problem andcreating all test cases

问题分析：一个简单的DFS（深度优先遍历）算法，这里避免开辟新空间来标志每个节点是否被访问，修改grid[][]对应的值不为题目设定的’0’或’1’即可；

代码：

public class Solution {
    public int numIslands(char[][] grid) {
		if(grid == null || grid.length == 0)
			return 0;	
		int result = 0;
		for(int i = 0; i < grid.length; i++)
		{
			for(int j = 0; j < grid[0].length; j++)
			{
				if(grid[i][j] == '1')
				{
					dfs_visit(grid, i, j);
					result ++;
				}
			}
		}
		return result;     
    }
	
	//访问函数
	private void dfs_visit(char[][] grid, int i, int j)
	{
		if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length)
			return;
		if(grid[i][j] != '1') 
			return;
		grid[i][j] = '2';//表示已经访问过该节点
		dfs_visit(grid, i + 1, j);
		dfs_visit(grid, i - 1, j);
		dfs_visit(grid, i, j + 1);
		dfs_visit(grid, i, j - 1);
		return;
	}
	
}