leetcode 200 : Number of Islands

最新推荐文章于 2025-06-12 16:17:46 发布

xudli

最新推荐文章于 2025-06-12 16:17:46 发布

阅读量1w

点赞数 4

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/xudli/article/details/45912547

leetcode 专栏收录该内容

216 篇文章

订阅专栏

本文介绍了一个算法问题——计算二维网格中岛屿的数量。通过递归搜索的方式标记相连的陆地为已访问，并计数独立的岛屿数量。适用于理解递归搜索和岛屿问题的解决方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Number of Islands

Total Accepted: 8305 Total Submissions: 38192

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:

Answer: 1

Example 2:

Answer: 3

[思路]

这道题有点像 word search的简化版. 每遇到'1'后, 开始向四个方向递归搜索. 搜到后变为'0', 因为相邻的属于一个island. 然后开始继续找下一个'1'.

[CODE]

public class Solution {
    public int numIslands(char[][] grid) {
        int count = 0;
        for(int i=0; i<grid.length; i++) {
            for(int j=0; j<grid[0].length; j++) {
                if(grid[i][j]=='1') {
                    search(grid, i, j);
                    ++count;
                }
            }
        }
        return count;
    }
    
    private void search(char[][] grid, int x, int y) {
        if(x<0 || x>=grid.length || y<0 || y>=grid[0].length || grid[x][y]!='1') return;
        grid[x][y] = '0';
        search(grid, x-1, y);
        search(grid, x+1, y);
        search(grid, x, y-1);
        search(grid, x, y+1);
    }
}