README_EN.md

841. Keys and Rooms

中文文档

Description

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.

Solutions

DFS.

Python3

class Solution:
    def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
        def dfs(u):
            if u in vis:
                return
            vis.add(u)
            for v in rooms[u]:
                dfs(v)

        vis = set()
        dfs(0)
        return len(vis) == len(rooms)

Java

class Solution {
    private List<List<Integer>> rooms;
    private Set<Integer> vis;

    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
        vis = new HashSet<>();
        this.rooms = rooms;
        dfs(0);
        return vis.size() == rooms.size();
    }

    private void dfs(int u) {
        if (vis.contains(u)) {
            return;
        }
        vis.add(u);
        for (int v : rooms.get(u)) {
            dfs(v);
        }
    }
}

C++

class Solution {
public:
    vector<vector<int>> rooms;
    unordered_set<int> vis;

    bool canVisitAllRooms(vector<vector<int>>& rooms) {
        vis.clear();
        this->rooms = rooms;
        dfs(0);
        return vis.size() == rooms.size();
    }

    void dfs(int u) {
        if (vis.count(u)) return;
        vis.insert(u);
        for (int v : rooms[u]) dfs(v);
    }
};

Go

func canVisitAllRooms(rooms [][]int) bool {
	vis := make(map[int]bool)
	var dfs func(u int)
	dfs = func(u int) {
		if vis[u] {
			return
		}
		vis[u] = true
		for _, v := range rooms[u] {
			dfs(v)
		}
	}
	dfs(0)
	return len(vis) == len(rooms)
}

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