Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considerred overlapping.

顾名思义，合并间隔

那么先排序，维护开头位置s，和递推当前最大能到的位置e

感觉和上一个题差不多

bool cmp(Interval x,Interval y){return (x.start<y.start);}
class Solution {
public:
    
    vector<Interval> merge(vector<Interval>& intervals) {
        int n=intervals.size();
        sort(intervals.begin(),intervals.end(),cmp);
        vector<Interval> ans;
        if(n==0)
            return ans;    
        int maxn=0,s=intervals[0].start,e=intervals[0].end;
        for(int i=1;i<n;i++){
            if(intervals[i].start<=e)
                e=max(e,intervals[i].end);
            else{
                ans.push_back(Interval(s,e));
                s=intervals[i].start;
                e=intervals[i].end;
            }
        }
        ans.push_back(Interval(s,e));
        return ans;
    }
};