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200. Number of Islands
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:
Input:
11110
11010
11000
00000
Output: 1Example 2:
Input:
11000
11000
00100
00011
Output: 3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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/**
* 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
}