Insert Interval

http://www.leetcode.com/onlinejudge

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].

/**Passed both small and large tests.
 * Be especially careful with the coundary cases
 * 
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (intervals.empty()) return vector<Interval>(1,newInterval);//base case 0
        int index1 = 0;
        for(; index1 < intervals.size(); ++index1){
            if (intervals[index1].start > newInterval.start) break;
        }
        if (index1){
            --index1;
            if (newInterval.start <= intervals[index1].end){//merge
                newInterval.start = intervals[index1].start;
            }else ++index1;//inclusive
        }
        int index2 = index1;//index2 >= 0
        for(; index2 < intervals.size(); ++index2){
            if (intervals[index2].end > newInterval.end) break;
        }
        if (index2 != intervals.size()){
            if (newInterval.end >= intervals[index2].start){
                newInterval.end = intervals[index2].end;
                ++index2;//skip the element
            }
        }
        vector<Interval> result(intervals.begin(),intervals.begin()+index1);
        result.push_back(newInterval);
        if (index2 <= intervals.size())
            result.insert(result.end(),intervals.begin()+index2, intervals.end());
        return result;
        
    }
};