There are N rooms and you start in room 0. Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room.

Formally, each room i has a list of keys rooms[i], and each key rooms[i][j] is an integer in [0, 1, ..., N-1] where N = rooms.length. A key rooms[i][j] = v opens the room with number v.

Initially, all the rooms start locked (except for room 0).

You can walk back and forth between rooms freely.

Return true if and only if you can enter every room.

Example 1:

Input: [[1],[2],[3],[]]
Output: true
Explanation:
We start in room 0, and pick up key 1.
We then go to room 1, and pick up key 2.
We then go to room 2, and pick up key 3.
We then go to room 3. Since we were able to go to every room, we return true.

Example 2:

Input: [[1,3],[3,0,1],[2],[0]]
Output: false
Explanation: We can't enter the room with number 2.

Note:

1 <= rooms.length <= 1000
0 <= rooms[i].length <= 1000
The number of keys in all rooms combined is at most 3000.

這個題的大意是首先在開始的門里面取鑰匙，然后依次打開鑰匙對應的門，最后看看門是否全部打開。

這個題可以用棧，也可以用隊列。在C++和Java有相應的庫可以調(diào)用。不過，在python中，可以用列表求解。

python代碼（基于隊列思維）:

class Solution:
    def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
        seen = [False] * len(rooms) # 類似于C++的memset(seen, false, sizeof(seen))。
        seen[0] = True
        q = [0] # 隊列
        
        while len(q):
            Q = q.pop(0)    # 彈出隊首元素。
            for index in rooms[Q]:
                if not seen[index]:
                    seen[index] = True
                    q.append(index)
        return all(seen)    # 判斷seen里面的元素是否都為True，即是否都打開了門。

在棧中，只需要在Q = q.pop(0)那個中改為Q = q.pop()可以了。