代码随想录图论 第五天| 841.钥匙和房间 463. 岛屿的周长

代码随想录图论 第五天| 841.钥匙和房间

一、 841.钥匙和房间

题目链接：https://leetcode.cn/problems/keys-and-rooms/
思路：钥匙就是索引，遍历过就标记，每拿到一个房间的钥匙，直接for循环递归遍历，深度优先直接拿下。

class Solution {
    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
        boolean[] visited = new boolean[rooms.size()];
        dfs(visited, rooms, 0);
        for (boolean b : visited) {
            if (b == false) return false;
        }
        return true;
    }
    void dfs(boolean[] visited, List<List<Integer>> rooms, int key) {
        if (visited[key]) return;
        visited[key] = true;
        List<Integer> list = rooms.get(key);
        for (Integer i : list) {
            dfs(visited, rooms, i);
        }
    }
}

二、463. 岛屿的周长

题目链接：https://leetcode.cn/problems/island-perimeter/
思路：常规思路，遍历每一个岛屿并且计算当前点的边长就可以。

class Solution {
    int sum = 0;
    public int islandPerimeter(int[][] grid) {
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    dfs(grid, i, j);
                    return sum;
                }
            }
        }
        return sum;
    }
    
    void dfs(int[][] grid, int x, int y) {
        if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) {
            return;
        }
        if (grid[x][y] == 2 || grid[x][y] == 0) return;
        grid[x][y] = 2;
        if (x-1 < 0 || grid[x-1][y] == 0) sum++;
        if (x+1 >= grid.length || grid[x+1][y] == 0) sum++;
        if (y-1 < 0 || grid[x][y-1] == 0) sum++;
        if (y+1 >= grid[0].length || grid[x][y+1] == 0) sum++;
        
        dfs(grid, x-1, y);
        dfs(grid, x+1, y);
        dfs(grid, x, y-1);
        dfs(grid, x, y+1);
    }
}