LeetCode--Keys and Rooms

Keys and Rooms更多技术博客 http://vilins.top/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: trueExplanation: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: falseExplanation: We can’t enter the room with number 2.Note:1 rooms.length 10000 rooms[i].length 1000The number of keys in all rooms combined is at most 3000.分析直接将问题转化为从一个起点0开始进行深度遍历看看是否有没遍历的顶点。源码class Solution { public: void dfs(vectorvectorint rooms, int index, vectorbool visited) { visited[index] true; for(int i 0; i rooms[index].size(); i) { if(!visited[rooms[index][i]]) { dfs(rooms, rooms[index][i], visited); } } } bool canVisitAllRooms(vectorvectorint rooms) { vectorbool visited(rooms.size(), false); dfs(rooms, 0, visited); for(int i 0; i rooms.size(); i) { if(!visited[i]) { return false; } } return true; } };更多技术博客 http://vilins.top/