435. Non-overlapping Intervals

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<Interval>& 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;
    }
};