[LeetCode] 841. Keys and Rooms [Java]

841. Keys and Rooms

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.

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각 방이 가지고 있는 열쇠들이 다음 방문할 수 있는 방을 뜻한다는 것만 이해하면, DFS를 이용해서 쉽게 풀 수 있다.
DFS를 통한 순회가 끝나면, visited[] 배열을 통해서 모든 방을 방문했는지 확인한다.

DFS임으로, 시간복잡도는 $$O(n+e)$$이다.

코드는 다음과 같다.

class Solution {
    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
        int n = rooms.size();
        boolean[] visited = new boolean[n];

        dfs(0, rooms, visited);

        for(boolean v : visited){
            if(!v){
                return false;
            }
        }

        return true;
    }

    public void dfs(int node, List<List<Integer>> rooms, boolean[] visited ){
        visited[node] = true;

        for(int next : rooms.get(node)){
            if(!visited[next]){
                dfs(next, rooms, visited);
            }
        }
    }

}