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].

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

struct Interval {
	int start;
	int end;
	Interval() : start(0), end(0) {}
	Interval(int s, int e) : start(s), end(e) {}
};

struct interval_less:public binary_function<const Interval,const Interval,bool>
{
	bool operator()(const Interval A,const Interval B){

		return A.start<B.start;
	}
};

vector<Interval> merge(vector<Interval> &intervals) {
	if (intervals.size()<=1)
		return intervals;
	vector<Interval> Result ;
	sort(intervals.begin(),intervals.end(),interval_less());
	Interval Cur_interval = *intervals.begin();
	vector<Interval>::iterator Nex_interval = intervals.begin()+1;
	while (Nex_interval!=intervals.end())
	{
		if ((*Nex_interval).start<=Cur_interval.end)
			Cur_interval = Interval(Cur_interval.start,max((*Nex_interval).end,Cur_interval.end));
		else{
			Result.push_back(Cur_interval);
			Cur_interval = *(Nex_interval);
		}
		Nex_interval++;
	}
	Result.push_back(Cur_interval);
	return Result;
}

Merge Intervals

原文：http://blog.csdn.net/li_chihang/article/details/44587637