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:

/** * 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 flag = true;        for(int i = 0; i < intervals.size(); ++i)        {            if(newInterval.end < intervals[i].start || newInterval.start > intervals[i].end)            {                res.push_back(intervals[i]);            }            else            {                int small = min(newInterval.start, intervals[i].start);                int big = max(newInterval.end, intervals[i].end);                if(flag) res.push_back(Interval(small, big));                else res[res.size()-1] = Interval(small, big);                newInterval = Interval(small, big);                flag = false;            }        }        if(flag)        {            if(intervals.size() == 0) res.push_back(newInterval);            else            {                int pos = 0;                while(pos < res.size() && newInterval.end > intervals[pos].start) pos++;                res.insert(res.begin()+pos, newInterval);            }        }        return res;    }};