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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 <= rooms.length <= 1000
0 <= rooms[i].length <= 1000
The number of keys in all rooms combined is at most 3000.
题目地址
讲解
这是一道典型的图的遍历，这里我采用了广度优先遍历，广度优先遍历可以用队列这种数据结构来实现。同时还需要一个 Set 用于记录已经访问过的 room。感觉这道题还是比较简单的
java 代码
class Solution {
public boolean canVisitAllRooms(List<List<Integer>> rooms) {
Set<Integer> set = new HashSet();
Queue<Integer> queue = new LinkedList<>();
for(Integer x:rooms.get(0)){
queue.offer(x);
}
while(!queue.isEmpty()){
int room = queue.poll();
if(set.contains(room)){
continue;
}else{
set.add(room);
List<Integer> list = rooms.get(room);
if(list!=null && list.size()>0){
for(Integer x: list){
queue.offer(x);
}
}
}
}
for(int i=1;i<rooms.size();i++){
if(!set.contains(i)){
return false;
}
}
return true;
}
}
