#### QUESTION:

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.

**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.
```

**Constraints:**

```
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.
```

#### EXPLANATION:

用一个keys来记录当前已经有的钥匙, 用一个visited来记录已经走过的房间, 在钥匙用光之后, 如果还没有走完全部房间, 就说明有房间走不了了. 为了避免重复, 已经走过的房间,也就是有钥匙的房间就不需要再走了.

#### SOLUTION:

```
class Solution {
func canVisitAllRooms(_ rooms: [[Int]]) -> Bool {
var keys:[Int] = rooms[0]
var visited = Set<Int>()
visited.insert(0)
while !keys.isEmpty {
var tmpKey = keys.removeFirst()
if !visited.contains(tmpKey) {
keys.append(contentsOf: rooms[tmpKey])
visited.insert(tmpKey)
}
}
return visited.count == rooms.count
}
}
```