문제

There are n rooms labeled from 0 to n - 1 and all the rooms are locked except for room 0. Your goal is to visit all the rooms. However, you cannot enter a locked room without having its key.

When you visit a room, you may find a set of distinct keys in it. Each key has a number on it, denoting which room it unlocks, and you can take all of them with you to unlock the other rooms.

Given an array rooms where rooms[i] is the set of keys that you can obtain if you visited room i, return true if you can visit all the rooms, or false otherwise.

[문제해석]

input: room array, set of room keys

output: room 방문할 때마다 True 반환, 다 방문할 수 있으면 False 반환

example 1:

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

example 2:

Input: rooms = [[1,3],[3,0,1],[2],[0]]
Output: false
Explanation: We can not enter room number 2 since the only key that unlocks it is in that room.

제한조건

n == rooms.length
2 <= n <= 1000
0 <= rooms[i].length <= 1000
1 <= sum(rooms[i].length) <= 3000
0 <= rooms[i][j] < n
All the values of rooms[i] are unique.

[풀이]

BFS(너비 우선 탐색

class Solution:
    def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:

        seen = set([False]) #set()을 씀으로써 봤던 곳을 또 보는 과정 생략
				stack = [0] #key 저장

				while stack:
					#큐 구현
					value = stack.pop()

					for i in rooms[value]:
						if i not in seen:
							seen.add(i)
	
							if len(seen) == len(rooms):
								return True
							stack.append(i)

				return len(stack) == len(rooms)

[배운점]

큐 구현하는 방법 ⇒ 여전히 어렵다.. 자료구조를 이용한 문제를 많이 풀어봐여 할 것 같음