結束的早不容易跟之後的overlap

class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        if(intervals.empty()) return 0;
        sort(intervals.begin(), intervals.end(), [](Interval& a, Interval& b){return a.start<b.start;});
        int pre_end = intervals[0].end;
        int res = 0;
        for(int i=1;i<intervals.size();i++){
            if(pre_end>intervals[i].start){
                res++;
                pre_end = min(pre_end, intervals[i].end);
            }else{
                pre_end = intervals[i].end;
            }
        }
        return res;
    }
};