README_EN.md

841. Keys and Rooms

中文文档

Description

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.

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)
}

TypeScript

function canVisitAllRooms(rooms: number[][]): boolean {
    const n = rooms.length;
    const isOpen = new Array(n).fill(false);
    const dfs = (i: number) => {
        if (isOpen[i]) {
            return;
        }
        isOpen[i] = true;
        rooms[i].forEach(k => dfs(k));
    };
    dfs(0);
    return isOpen.every(v => v);
}

function canVisitAllRooms(rooms: number[][]): boolean {
    const n = rooms.length;
    const isOpen = new Array(n).fill(false);
    const keys = [0];
    while (keys.length !== 0) {
        const i = keys.pop();
        if (isOpen[i]) {
            continue;
        }
        isOpen[i] = true;
        keys.push(...rooms[i]);
    }
    return isOpen.every(v => v);
}

Rust

impl Solution {
    pub fn can_visit_all_rooms(rooms: Vec<Vec<i32>>) -> bool {
        let n = rooms.len();
        let mut is_open = vec![false; n];
        let mut keys = vec![0];
        while !keys.is_empty() {
            let i = keys.pop().unwrap();
            if is_open[i] {
                continue;
            }
            is_open[i] = true;
            rooms[i].iter().for_each(|&key| keys.push(key as usize));
        }
        is_open.iter().all(|&v| v)
    }
}

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