Insert intervals

题目：

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].

思路： 对于intervals从头到尾遍历，用new interval与相应的interval[i]对比，如果整个new interval都在interval[i]的左边，就把new interval存到result里，然后把剩下的所有interval都存起来即可，如果new interval在interval[i]的右边，则存起来interval[i]，然后继续遍历； 如果new interval 和 interval[i]有overlap， 则将两者进行合并，然后继续遍历。

代码： 

class Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<Interval> result;
        
        for (int i = 0; i<intervals.size(); i++){
            
            if (intervals[i].end < newInterval.start)
               result.push_back(intervals[i]);
            else if (intervals[i].start>newInterval.end){
                
                result.push_back(newInterval);
                while (i<intervals.size()){
                    
                    result.push_back(intervals[i]);
                    i++;
                }
            }
            
            else{
                
                newInterval.start = min(intervals[i].start, newInterval.start);
                newInterval.end = max(intervals[i].end, newInterval.end);
                
            }
        }
        
        if(0 == result.size() || result.back().end < newInterval.start)  
            result.push_back(newInterval);  
        
        return result;
    }
};