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200. 岛屿数量
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。
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：
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m == grid.length -
n == grid[i].length -
1 <= m, n <= 300 -
grid[i][j]的值为0或1
思路分析
思考题：看题解可以使用 UnionFind 来解决这个问题。可以思考一下，如何实现？
这道题与 0130-surrounded-regions.adoc 类似。
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一刷
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二刷
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三刷
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/**
* Runtime: 1 ms, faster than 99.99% of Java online submissions for Number of Islands.
*
* Memory Usage: 42 MB, less than 41.86% of Java online submissions for Number of Islands.
*
* Copy from: https://leetcode-cn.com/problems/number-of-islands/solution/dao-yu-shu-liang-by-leetcode/[岛屿数量 - 岛屿数量 - 力扣（LeetCode）]
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-01-25 23:42
*/
public int numIslands(char[][] grid) {
if (Objects.isNull(grid) || grid.length == 0) {
return 0;
}
int yLength = grid.length;
int xLength = grid[0].length;
int result = 0;
for (int y = 0; y < yLength; y++) {
for (int x = 0; x < xLength; x++) {
if (grid[y][x] == '1') {
result++;
dfs(grid, y, x);
}
}
}
return result;
}
private void dfs(char[][] grid, int y, int x) {
int yLength = grid.length;
int xLength = grid[0].length;
if (y < 0 || y >= yLength || x < 0 || x >= xLength
|| grid[y][x] == '0') {
return;
}
grid[y][x] = '0';
dfs(grid, y - 1, x);
dfs(grid, y + 1, x);
dfs(grid, y, x - 1);
dfs(grid, y, x + 1);
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-11 20:04:34
*/
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int result = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
if (grid[i][j] == '1') {
result++;
dfs(grid, i, j);
}
}
}
return result;
}
private void dfs(char[][] grid, int row, int col) {
if (row < 0 || grid.length <= row
|| col < 0 || grid[0].length <= col) {
return;
}
if (grid[row][col] == '0' || grid[row][col] != '1') {
return;
}
grid[row][col] = '2';
dfs(grid, row - 1, col);// 上
dfs(grid, row + 1, col);// 下
dfs(grid, row, col - 1); // 左
dfs(grid, row, col + 1); // 右
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2025-04-06 15:18:59
*/
public int numIslands(char[][] grid) {
int result = 0;
for (int r = 0; r < grid.length; r++) {
for (int c = 0; c < grid[r].length; c++) {
if (grid[r][c] == '1') {
result++;
dfs(grid, r, c);
}
}
}
return result;
}
private void dfs(char[][] grid, int r, int c) {
if (r < 0 || grid.length <= r
|| c < 0 || grid[r].length <= c
|| grid[r][c] != '1') {
return;
}
grid[r][c] = '2';
dfs(grid, r - 1, c);
dfs(grid, r + 1, c);
dfs(grid, r, c - 1);
dfs(grid, r, c + 1);
}