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/
-
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)
广度优先搜索：类似的操作
并查集：新概念，没太理解
