Last Updated on 2022-12-20 by Clay
題目
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.length2 <= n <= 10000 <= rooms[i].length <= 10001 <= sum(rooms[i].length) <= 30000 <= rooms[i][j] < n- All the values of
rooms[i]are unique.
題目會給我們許多房間，每個房間內都有對應特定門的鑰匙。我們最一開始會擁有第 0 號房間的鑰匙，最後要判斷我們是否能走完所有的房間。
解題思路
今天工作較忙，只附上範例程式。
C++ 是 DFS、Python 是某種程度上的暴力解。
C++ 範例程式碼
class Solution {
public:
void DFS(vector<vector<int>>& rooms, unordered_map<int, bool>& visited, int n) {
for (auto& key: rooms[n]) {
if (!visited[key]) {
visited[key] = true;
DFS(rooms, visited, key);
}
}
}
bool canVisitAllRooms(vector<vector<int>>& rooms) {
unordered_map<int, bool> visited({{0, true}});
DFS(rooms, visited, 0);
return visited.size() == rooms.size();
}
};Python 範例程式碼
class Solution:
def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
setA = set([0])
setB = set()
setC = set([0])
while setC:
for n in setC:
setB.update(set(rooms[n]))
setC = setB.difference(setA)
setA.update(setB)
setB.clear()
return len(setA) == len(rooms)