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].

既然有序了，就简单了。

class Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        int i = 0;
        int len = intervals.size();
        if(0 == len) { intervals.push_back(newInterval); return intervals;}
        vector<Interval> re;
        //find the insert position of start.
        while(i < len && intervals[i].end < newInterval.start){
            re.push_back(intervals[i]);
            ++i;
        }
        //at the right, 
        if(i == len){re.push_back(newInterval); return re; }
        int j = i;
        //find the insert position of end.
        while(j < len && intervals[j].end < newInterval.end){
            ++j;
        }
        Interval jion;
        jion.start = min(intervals[i].start, newInterval.start);
        if(j == len){
            jion.end = newInterval.end;
            re.push_back(jion);
            return re;
        }
    
        if(newInterval.end < intervals[j].start){
           --j;
           jion.end = newInterval.end;
           re.push_back(jion);
        } 
        else{
            jion.end = intervals[j].end;
            re.push_back(jion);
        }
        ++j;
        while(j < len){
            re.push_back(intervals[j]);
            ++j;
        }
        return re;
    }
};