Keys and Rooms

There areNrooms and you start in room0. Each room has a distinct number in0, 1, 2, ..., N-1, and each room may have some keys to access the next room.

Formally, each roomi has a list of keysrooms[i], and each keyrooms[i][j]is an integer in[0, 1, ..., N-1]whereN = rooms.length. A keyrooms[i][j] = v opens the room with numberv.

Initially, all the rooms start locked (except for room0).

You can walk back and forth between rooms freely.

Returntrue if and only if you can enter every room.

1. 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.

分析

除了初始入栈visited也入0，后面是出栈时入visited

class Solution:
    def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
        q = []+rooms[0]
        v = set()
        v.add(0)
        while q:
            cur = q.pop()
            if cur in v:
                continue
            v.add(cur)

            for u in rooms[cur]:
                if u not in v:
                    q.append(u)

        return len(v) == len(rooms)

DFS

起始点开始，每个房间走下去

class Solution:
    def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:        
        v = set()
        def dfs(i):
            if i in v:
                return
            v.add(i)
            for j in rooms[i]:
                dfs(j)

        dfs(0)            
        return len(v) == len(rooms)