Insert Interval

题目描述：

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].

解题思路：

由于先前的间隔区间数组已经是有序的，所以首先通过插入排序把新加入的区间插入到区间数组中，
然后一趟遍历，合并存在重叠的间隔区间即可

AC代码如下：

class Solution {
public:
	vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
		vector<Interval> ans;
		if (intervals.size() == 0){
			ans.push_back(newInterval);
			return ans;
		}
		intervals.push_back(newInterval);
		int i = intervals.size() - 2;
		for (; i >= 0; --i){
			if (newInterval.start < intervals[i].start){
				intervals[i + 1] = intervals[i];
			}else{
			    break;
			}
		}
		intervals[i + 1] = newInterval;
		Interval cur = intervals[0];
		for (int j = 1; j < intervals.size(); ++j){
			if (intervals[j].start >= cur.start && intervals[j].start <= cur.end){
				cur = Interval(cur.start, max(cur.end, intervals[j].end));
			}
			else{
				ans.push_back(cur);
				cur = intervals[j];
			}
		}
		ans.push_back(cur);
		return ans;
	}
};