LeetCode_DFS_中等_200.岛屿数量

￡神魔★判官ぃ 2023-10-06 22:29 1阅读 0赞

#### 目录 ####

*  1.题目
 *  2.思路
 *  3.代码实现（Java）

## 1.题目 ##

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’

来源：力扣（LeetCode）  
链接：https://leetcode-cn.com/problems/number-of-islands

## 2.思路 ##

（1）DFS  
思路参考[一文秒杀所有岛屿题目][Link 1]。

相关题目：  
[LeetCode\_DFS\_中等\_面试题 16.19.水域大小][LeetCode_DFS_ 16.19.]

## 3.代码实现（Java） ##

//思路1————DFS
    class Solution {
        
        public int numIslands(char[][] grid) {
        
            // res 保存岛屿数量，初始值为 0
            int res = 0;
            int m = grid.length;
            int n = grid[0].length;
            //遍历 grid
            for (int i = 0; i < m; i++) {
        
                for (int j = 0; j < n; j++) {
        
                    if (grid[i][j] == '1') {
        
                        //发现一个岛屿
                        res++;
                        //使用 DFS 将该岛屿淹没(包括上下左右相邻的陆地)
                        dfs(grid, i, j);
                    }
                }
            }
            return res;
        }
    
        public void dfs(char[][] grid, int i, int j) {
        
            int m = grid.length;
            int n = grid[0].length;
            //判断边界条件
            if (i < 0 || j < 0 || i >= m || j >= n) {
        
                return;
            }
            if (grid[i][j] == '0') {
        
                //当前位置已经是海水，则直接返回
                return;
            }
            //将当前位置变成海水，相当于使用数组 visited 记录遍历过的节点
            grid[i][j] = '0';
            //同时也淹没上下左右的陆地
            dfs(grid, i - 1, j);
            dfs(grid, i + 1, j);
            dfs(grid, i, j - 1);
            dfs(grid, i, j + 1);
        }
    }

[Link 1]: https://labuladong.github.io/algo/4/29/108/
[LeetCode_DFS_ 16.19.]: https://blog.csdn.net/weixin_43004044/article/details/131338530