leetcode
200. 岛屿数量

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''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
    }
}