Insert Interval

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].

Solution: For example 2:
1. compare [1,2] with [4,9], then insert [1,2];
2. merge [3,5] with [4,9], get newInterval = [3,9];
3. merge [6,7] with [3,9], get newInterval = [3,9];
4. merge [8,10] with [3,9], get newInterval = [3,10];
5. compare [12,16] with [3,10], insert newInterval [3,10], then all the remaining intervals...

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        vector<Interval> res;
        bool insert = false;
        for(vector<Interval>::iterator it = intervals.begin(); it != intervals.end(); it++) {
            if(insert || it->end < newInterval.start) {
                res.push_back(*it);
            }
            else if(newInterval.end < it->start) {
                res.push_back(newInterval);
                res.push_back(*it);
                insert = true;
            }
            else {
                newInterval.start = min(newInterval.start, it->start);
                newInterval.end = max(newInterval.end, it->end);
            }
        }
        if(!insert) res.push_back(newInterval);
        return res;
    }
};