Number of Islands

最新推荐文章于 2025-05-07 17:27:01 发布

原创最新推荐文章于 2025-05-07 17:27:01 发布 · 367 阅读

0 ·

CC 4.0 BY-SA版权

strStr 同时被 3 个专栏收录

360 篇文章

订阅专栏

开发智力

67 篇文章

订阅专栏

候着

26 篇文章

订阅专栏

本文介绍了一种使用深度优先搜索（DFS）算法解决岛屿数量计算问题的方法。通过遍历二维网格地图，将每个‘1’（陆地）标记为‘0’（水域），以此消除相连的陆地并计算岛屿总数。

新知，扫描整个grid，将1的周围都设为0（dfs），这样有几个1，就有几个岛

至于用union-find的解法，之后研习

好了，可以先去吃午饭了

public class Solution {
    public int numIslands(char[][] grid) {
        int num = 0;
        if (grid == null || grid.length == 0 || grid[0] == null || grid[0].length == 0) {
            return 0;
        }
        for (int x = 0; x < grid.length; x++) {
            for (int y = 0; y < grid[0].length; y++) {
                if (grid[x][y] == '1') {
                    num++;
                    numIslandsHelper(x, y, grid);
                }
            }
        }
        return num;
    }
    
    private void numIslandsHelper(int x, int y, char[][] grid) {
        grid[x][y] = '0';
        if (x - 1 >= 0 &&  grid[x - 1][y] == '1') {
            numIslandsHelper(x - 1, y, grid);
        }
        if (y - 1 >= 0 &&  grid[x][y - 1] == '1') {
            numIslandsHelper(x, y - 1, grid);
        }
        if (x + 1 < grid.length &&  grid[x + 1][y] == '1') {
            numIslandsHelper(x + 1, y, grid);
        }
        if (y + 1 < grid[0].length &&  grid[x][y + 1] == '1') {
            numIslandsHelper(x, y + 1, grid);
        }
    }
}

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: