给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。此外，你可以假设该网格的四条边均被水包围。

https://leetcode-cn.com/problems/number-of-islands/

示例1：

输入：grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出：1

示例2：

输入：grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出：3

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

Java解法

思路：

• 这道题与做过的Day67 被围绕的区域类似，可以用相同的方式处理
• 从开始位置进行遍历，遇到非标记陆地时进行计数，同时以该陆地为起点做链接陆地的查找，对该值进行标记
• 遍历结束，返回计数

package sj.shimmer.algorithm.m4_2021;

/**
 * Created by SJ on 2021/4/18.
 */

class D81 {
    public static void main(String[] args) {
        System.out.println(numIslands(new char[][]{
                {'1', '1', '1', '1', '0'},
                {'1', '1', '0', '1', '0'},
                {'1', '1', '0', '0', '0'},
                {'0', '0', '0', '0', '0'}
        }));
        System.out.println(numIslands(new char[][]{
                {'1','1','0','0','0'},
                {'1','1','0','0','0'},
                {'0','0','1','0','0'},
                {'0','0','0','1','1'}
        }));
    }
    public static int numIslands(char[][] grid) {
        int count = 0;
        if (grid != null && grid.length > 0 && grid[0] != null) {
            int height = grid.length;
            int width = grid[0].length;
            for (int w = 0; w < width; w++) {
                for (int h = 0; h < height; h++) {
                    if (grid[h][w] == '1') {
                        signLand(grid, w, h);
                        count++;
                    }
                }
            }
        }
        return count;
    }

    public static void signLand(char[][] grid, int w, int h) {
        if (w >= 0 && w < grid[0].length && h < grid.length && h >= 0) {
            if (grid[h][w] == '1') {
                grid[h][w] = '2';
                signLand(grid, w - 1, h);
                signLand(grid, w, h - 1);
                signLand(grid, w + 1, h);
                signLand(grid, w, h + 1);
            }
        }
    }
}

image

官方解

https://leetcode-cn.com/problems/number-of-islands/solution/dao-yu-shu-liang-by-leetcode/

1. 深度优先搜索

   类似于我的解法，但有无向图的概念，我没具体了解

   • 时间复杂度：O(MN)
   • 空间复杂度：O(MN)

2. 广度优先搜索：类似的操作

3. 并查集：新概念，没太理解