对每个询问，先找出符合实际房间体积，大于询问房间体积房间的房间号，之后再从这些符合条件的候选答案中挑选一个最接近询问房间ID号的。
为了找出符合条件的房间，可以对实际房间、询问房间，按照房间体积降序排序，然后只将大于等于询问房间体积的房间ID，加入到待筛选的房间列表中。因为对于小于询问房间体积的房间，是不会作为答案的。
同时保证待筛选的房间列表保存在一个有序的列表之中，这样就可以通过二分查找的方式，快速的找到最接近询问房间ID的房间ID，实现如下：
struct record
{int originId;int roomSize;int roomId;record(int _originId, int _roomSize, int _roomId){this->originId = _originId;this->roomSize = _roomSize;this->roomId = _roomId;}
};bool compare__(const record& a, const record& b)
{return a.roomSize > b.roomSize;
}class Solution {
public:vector<int> closestRoom(vector<vector<int>>& rooms, vector<vector<int>>& queries) {vector<record> roomRecords;vector<record> queryRecords;for(int i = 0; i < rooms.size(); i++){roomRecords.emplace_back(record(i, rooms[i][1], rooms[i][0]));}for(int i = 0; i < queries.size(); i++){queryRecords.emplace_back(record(i, queries[i][1], queries[i][0]));}sort(roomRecords.begin(), roomRecords.end(), compare__);sort(queryRecords.begin(), queryRecords.end(), compare__);vector<int> res(queries.size(), -1);set<int> curRoomIdSet;int j = 0;for(int i = 0; i < queryRecords.size(); i++){int curRoomID = queryRecords[i].roomId;int curRoomSize = queryRecords[i].roomSize;int curQueryID = queryRecords[i].originId;while(j < roomRecords.size() && roomRecords[j].roomSize >= curRoomSize){curRoomIdSet.insert(roomRecords[j].roomId);j++;}int dist = INT_MAX;auto it = curRoomIdSet.lower_bound(curRoomID);if(it != curRoomIdSet.begin()){auto p = prev(it);dist = curRoomID - *p;res[curQueryID] = *p;}if(it != curRoomIdSet.end() && *it - curRoomID < dist)res[curQueryID] = *it;}return res;}
};
