问题描述：
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:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]Example 2:
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
题源：here；完整实现：here
思路：
我们使用二分查找的方式找到newInterval.start和newInterval.end的位置，然后删除中间的区间；之后插入newInterval并重新检查，代码如下：
class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
if (intervals.size() == 0) return{ newInterval };
int startPos, endPos;
int left = 0, right = intervals.size() - 1;
while (left < right){
int mid = (left + right + 1) / 2;
if (intervals[mid].start <= newInterval.start) left = mid;
else right = mid - 1;
}
startPos = left;
left = 0; right = intervals.size() - 1;
while (left < right){
int mid = (left + right) / 2;
if (intervals[mid].end <= newInterval.end) left = mid + 1;
else right = mid;
}
endPos = right;
if (endPos - startPos == 2) intervals.erase(intervals.begin() + startPos + 1);
else if (endPos - startPos > 2) intervals.erase(intervals.begin() + startPos + 1, intervals.begin() + endPos);
if (intervals[startPos].start>newInterval.start) intervals.insert(intervals.begin() + startPos, newInterval);
else intervals.insert(intervals.begin() + startPos + 1, newInterval);
int idx = startPos;
for (int i = startPos; i < startPos + 2 && idx<intervals.size() - 1; i++){
if (intervals[idx].end >= intervals[idx + 1].start){
intervals[idx].end = max(intervals[idx].end, intervals[idx + 1].end);
intervals.erase(intervals.begin() + idx + 1);
}
else idx++;
}
return intervals;
}
};