力扣算法 Java 刷题笔记【DFS篇】hot100（三）岛屿问题（一） 3

文章目录

• 1. 岛屿数量（中等）
• 2. 统计封闭岛屿的数目（中等）
• 3. 飞地的数量（中等）

1. 岛屿数量（中等）

地址: https://leetcode-cn.com/problems/number-of-islands/
2022/01/27
做题反思：

class Solution {
    public int numIslands(char[][] grid) {
        int res = 0;
        int m = grid.length, n = grid[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    res++;
                    dfs(grid, i, j);
                }
            }
        }
        return res;
    }
    void dfs(char[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length) {
            return;
        }
        if (grid[i][j] == '0') {
            return;
        }
        grid[i][j] = '0';
        dfs(grid, i, j + 1);
        dfs(grid, i, j - 1);
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
    }
}

2. 统计封闭岛屿的数目（中等）

地址: https://leetcode-cn.com/problems/number-of-closed-islands/
2022/01/27
做题反思：

class Solution {
    public int closedIsland(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        for (int i = 0; i < m; i++) {
            dfs(grid, i, 0);
            dfs(grid, i, n - 1);
        }
        for (int j = 0; j < n; j++) {
            dfs(grid, 0, j);
            dfs(grid, m - 1, j);
        }
        int res = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0) {
                    res++;
                    dfs(grid, i , j);
                }
            }
        }
        return res;
    }

    void dfs(int[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length) {
            return;
        }
        if (grid[i][j] == 1) {
            return;
        }
        grid[i][j] = 1;
        dfs(grid, i, j + 1);
        dfs(grid, i, j - 1);
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
    }
}

3. 飞地的数量（中等）

地址: https://leetcode-cn.com/problems/number-of-enclaves/
2022/01/27
做题反思：

class Solution {
    public int numEnclaves(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        for (int i = 0; i < m; i++) {
            dfs(grid, i, 0);
            dfs(grid, i, n - 1);
        }
        for (int j = 0; j < n; j++) {
            dfs(grid, 0, j);
            dfs(grid, m - 1, j);
        }
        int res = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    res++;
                }
            }
        }
        return res;
    }

    void dfs(int[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length) {
            return;
        }
        if (grid[i][j] == 0) {
            return;
        }
        grid[i][j] = 0;
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
    }
}