https://leetcode.com/problems/non-overlapping-intervals/description/

Greedy. Sort by (start,-end). Scan from left to right, if we see 2 intervals overlap, we should remove the interval with larger end value.

/** * Definition for an interval. * struct Interval { *     int start; *     int end; *     Interval() : start(0), end(0) {} *     Interval(int s, int e) : start(s), end(e) {} * }; */class Solution {public:    int eraseOverlapIntervals(vector
     
      & intervals) {        if (intervals.size() <= 1)  return 0;                auto comp = [](const Interval& i1, const Interval& i2) {            return i1.start < i2.start || i1.start == i2.start && i1.end > i2.end;        };        sort(intervals.begin(), intervals.end(), comp);                int res = 0;        auto cur = intervals[0];        for (int i = 1; i < intervals.size(); i++) {            if (cur.end > intervals[i].start) {                res++;                if (cur.end > intervals[i].end) {   // delete cur                    cur = intervals[i];                }            }            else {  // no overlap                cur = intervals[i];            }        }        return res;    }};