[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;

    }

};