[LeetCode]Merge Intervals

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].

思考：先排序。比较相邻两个interval，left为前一个start，若后一个start小于等于前一个，说明这两个interval需要合并，right为两个end较大者。若后一个start大于前一个end，说明这两个interval不需要合并，输出前一个。

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
bool comp(const Interval &a,const Interval &b)
{
	return a.start<b.start;
}
class Solution {
private:
	vector<Interval> ret;
public:

		vector<Interval> merge(vector<Interval> &intervals) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		ret.clear();
        int len=intervals.size();
		if(len==0) return ret;
		if(len==1) {ret.push_back(intervals[0]);return ret;}
		sort(intervals.begin(),intervals.end(),comp);
		int left=intervals[0].start;
		int right=intervals[0].end;
		for(int i=1;i<len;i++)
		{
			if(intervals[i].start<=right)
				right=max(right,intervals[i].end);
			if(intervals[i].start>right)
			{
		//		cout<<left<<" "<<right<<endl;
				ret.push_back(Interval(left,right));
				left=intervals[i].start;
				right=intervals[i].end;
			}
			if(i==len-1){
		//	    cout<<left<<" "<<right<<endl;
			ret.push_back(Interval(left,right));}
		}
		return ret;
    }
};