问题描述
LeetCode 200. 岛屿数量 (opens in a new tab),难度中等。
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:
m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j]
的值为'0'
或'1'
题解
Solution.java
class Solution {
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0) return 0;
int ans = 0;
int rows = grid.length;
int cols = grid[0].length;
for (int r = 0; r < rows; r++) {
for (int c = 0; c < cols; c++) {
if (grid[r][c] == '1') {
ans++;
// 将坐标 (r, c) 周围的陆地都设置为 0
dfs(grid, r, c);
}
}
}
return ans;
}
private void dfs(char[][] grid, int r, int c) {
int rows = grid.length;
int cols = grid[0].length;
if (r < 0 || c < 0 || r >= rows || c >= cols || grid[r][c] == '0') {
return;
}
grid[r][c] = '0';
dfs(grid, r - 1, c); // 上
dfs(grid, r + 1, c); // 下
dfs(grid, r, c - 1); // 左
dfs(grid, r, c + 1); // 右
}
}