Given a collection of intervals, merge all overlapping intervals.

For example,

Given [1,3],[2,6],[8,10],[15,18],

return [1,6],[8,10],[15,18].

bool operator < (const Interval &a, const Interval &b){ return a.start < b.start;}class Solution {public:    std::vector
     
       merge(std::vector
      
        &intervals) {        if(intervals.size() == 0) return intervals;		std::vector
       
         res;		std::sort(intervals.begin(),intervals.end());		int left = intervals[0].start,right = intervals[0].end;		for (int i = 1; i < intervals.size(); i++)		{			if(intervals[i].start <= right)			{				right = std::max(right,intervals[i].end);			}			else			{				res.push_back(Interval(left,right));				left = intervals[i].start;				right = intervals[i].end;			}		}		res.push_back(Interval(left,right));		return res;    }};
       
      
     

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