算法-岛屿数量

沉淀19582024-03-08 5:08

1、题目来源

200. 岛屿数量 - 力扣（LeetCode）

2、题目描述

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：

复制代码

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：

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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'

3、题解分享

java 复制代码

class Solution {
    public int numIslands(char[][] grid) {
        // 思路：深度遍历 + 修改遍历过的位置
        int n = grid.length;
        int m = grid[0].length;
        int ans = 0;
        for(int i=0;i<n;++i){
            for(int j = 0;j<m;++j){
                if(grid[i][j] == '1'){//每遇到一块岛屿
                    ++ans;
                    dfs(grid,n,m,i,j);
                }
            }
        }
        return ans;
    }
    public void dfs(char[][] grid,int n,int m,int row,int col){
        if(row <0 || row >= n || col < 0 || col >=m){
            return;
        }
        if(grid[row][col] == '0'){
            return;
        }

        grid[row][col] = '0';

        dfs(grid,n,m,row-1,col);
        dfs(grid,n,m,row+1,col);
        dfs(grid,n,m,row,col-1);
        dfs(grid,n,m,row,col+1);
    }
}