【LeetCode】841.钥匙和房间（bfs+dfs，java实现）

题目

链接

题解

方法一：bfs

class Solution {
    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
        boolean[] seen = new boolean[rooms.size()];
        seen[0] = true;
        Stack<Integer> stack = new Stack();
        stack.push(0);

        //At the beginning, we have a todo list "stack" of keys to use.
        //'seen' represents at some point we have entered this room.
        while (!stack.isEmpty()) { // While we have keys...
            int node = stack.pop(); // Get the next key 'node'
            for (int nei: rooms.get(node)) // For every key in room # 'node'...
                if (!seen[nei]) { // ...that hasn't been used yet
                    seen[nei] = true; // mark that we've entered the room
                    stack.push(nei); // add the key to the todo list
                }
        }

        for (boolean v: seen)  // if any room hasn't been visited, return false
            if (!v) return false;
        return true;
    }
}

方法二：dfs

class Solution {
    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
        boolean[] visit = new boolean[rooms.size()];
        dfs(rooms,0,visit);
        for (boolean visited:visit){
            if (!visited)
                return false;
        }
        return true;
    }

    private void dfs(List<List<Integer>> rooms, int room, boolean[] visit){
        if (!visit[room]){
            visit[room]=true;
            for (int num:rooms.get(room))
                dfs(rooms,num,visit);
        }
    }
}