力扣1942.最小未被占据椅子的编号
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用两个对组数组存到达的离开时间及其编号
- 遍历到达时间 同时处理之前所有离开时间
- 离开就把座位清空 再给当前到达的安排座位
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class Solution {public:int smallestChair(vector<vector<int>>& times, int targetFriend) {int n = times.size();vector<pair<int,int>> arrival;vector<pair<int,int>> leaving;for(int i=0;i<n;i++){arrival.emplace_back(times[i][0],i);leaving.emplace_back(times[i][1],i);}sort(arrival.begin(), arrival.end());sort(leaving.begin(), leaving.end());priority_queue<int,vector<int>,greater<int>> available;for(int i=0;i<n;i++)available.push(i);unordered_map<int,int> seats;int j = 0;for(auto &&[time,person] : arrival){while(j < n && leaving[j].first <= time){available.push(seats[leaving[j].second]);j ++;}int seat = available.top();seats[person] = seat;available.pop();if(person == targetFriend)return seat;}return -1;}};
)
