0841. Keys and Rooms

# 841. Keys and Rooms#

## 题目 #

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`.

## 题目大意 #

• 1 <= rooms.length <= 1000
• 0 <= rooms[i].length <= 1000
• 所有房间中的钥匙数量总计不超过 3000。

## 解题思路 #

• 给出一个房间数组，每个房间里面装了一些钥匙。0 号房间默认是可以进入的，房间进入顺序没有要求，问最终能否进入所有房间。
• 用 DFS 依次深搜所有房间的钥匙，如果都能访问到，最终输出 true。这题算是 DFS 里面的简单题。

## 代码 #

``````func canVisitAllRooms(rooms [][]int) bool {
visited := make(map[int]bool)
visited[0] = true
dfsVisitAllRooms(rooms, visited, 0)
return len(rooms) == len(visited)
}

func dfsVisitAllRooms(es [][]int, visited map[int]bool, from int) {
for _, to := range es[from] {
if visited[to] {
continue
}
visited[to] = true
dfsVisitAllRooms(es, visited, to)
}
}
``````

Sep 6, 2020