leetcode

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
struct Interval
{
	int start;
	int end;
	Interval() : start(0), end(0) {}
	Interval(int s, int e) : start(s), end(e) {}
};
bool operator <(const Interval a, const Interval b) { return a.start < b.start; }
class Solution {
public:
    std::vector<Interval> insert(std::vector<Interval> &intervals, Interval newInterval) {
		std::vector<Interval> res;
		intervals.push_back(newInterval);
		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;
    }
};