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;
}
};