LeetCode 435. Non-overlapping Intervals

贪心

按照有边界排序，只有先选择了右边边界小的，才可以放下更多的区间

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