LeetCode - 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]

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
 bool cmp(const Interval &a, const Interval &b){
    return a.start<=b.start;
 }
class Solution {
public:
    int merge(Interval &a,Interval &b){
		int &s1=a.start,&s2=b.start,&e1=a.end,&e2=b.end;
		if(s1>s2){//b.start在a.start前
			if(e2<s1){//相离
				return 0;
			}
			if(s1<=e2&&e2<=e1){//相交
				e2=e1;
				return 2;
			}
			if(e1<e2){//包含
				return 2;
			} 
		}
	
		if(e1<s2){//相离
			return 1;
		}
		if(s2<=e1&&e1<=e2){//相交
			s2=s1;
			return 2;
		}
		if(e2<e1){//包含
			s2=s1;
			e2=e1;
			return 2;
		}
    }

    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
     	vector<Interval> result;
		if(intervals.empty()||intervals.size()==0){
			result.push_back(newInterval);
			return result;
		}
		sort(intervals.begin(),intervals.end(),cmp);
		int i=0;
		bool inserted=false;
		for(;i<intervals.size();i++){
			if(merge(intervals[i],newInterval)==0){
				result.push_back(newInterval);
				inserted=true;
				break;
			}else if(merge(intervals[i],newInterval)==1){
				result.push_back(intervals[i]);
			}
		}
		for(;i<intervals.size();i++){
				result.push_back(intervals[i]);
		}
		if(!inserted){
			result.push_back(newInterval);
		}
		return result;
        
    }
};