# 841. Keys and Rooms 钥匙和房间

@TOC

## # 题目描述

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. 1 <= rooms.length <= 1000
2. 0 <= rooms[i].length <= 1000
3. The number of keys in all rooms combined is at most 3000.

## # 解题方法

### # DFS

``````class Solution:
def canVisitAllRooms(self, rooms):
"""
:type rooms: List[List[int]]
:rtype: bool
"""
visited = [0] * len(rooms)
self.dfs(rooms, 0, visited)
return sum(visited) == len(rooms)

def dfs(self, rooms, index, visited):
visited[index] = 1
for key in rooms[index]:
if not visited[key]:
self.dfs(rooms, key, visited)
``````

C++代码如下：

``````class Solution {
public:
bool canVisitAllRooms(vector<vector<int>>& rooms) {
int N = rooms.size();
vector<int> visited(N);
dfs(visited, rooms, 0);
int res = 0;
for (int v : visited) res += v;
return res == N;
}
private:
void dfs(vector<int>& visited, vector<vector<int>>& rooms, int pos) {
visited[pos] = 1;
for (int n : rooms[pos])
if (!visited[n])
dfs(visited, rooms, n);
}
};
``````

### # BFS

``````class Solution {
public:
bool canVisitAllRooms(vector<vector<int>>& rooms) {
int N = rooms.size();
vector<int> visited(N);
queue<int> q;
q.push(0);
while (!q.empty()) {
int f = q.front(); q.pop();
if (visited[f]) continue;
visited[f] = 1;
for (int n : rooms[f]) {
q.push(n);
}
}
int res = 0;
for (int v : visited) res += v;
return res == N;
}
};
``````

## # 日期

2018 年 5 月 28 日 —— 太阳真的像日光灯～ 2018 年 12 月 6 日 —— 周四啦！