leetcode[57]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) {}
 * };
 */
class Solution {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) 
{
    vector<Interval> res;
    Interval temp(newInterval.start,newInterval.end);
    for (vector<Interval>::iterator iter=intervals.begin();iter!=intervals.end();iter++)
    {
        if (temp.end<iter->start)
        {
            res.push_back(temp);
            res.insert(res.end(),iter,intervals.end());
            return res;
        }
        else if (temp.start>iter->end)
        {
            res.push_back(*iter);
        }
        else
        {
            temp.start=temp.start<=iter->start?temp.start:iter->start;
            temp.end=temp.end>=iter->end?temp.end:iter->end;
        }
    }
    res.push_back(temp);
    return res;
}
};

/**
采用自定义的比较函数，必须申明为全局对象，不能是类的局部对象。
 */
 /**
 bool comp(const Interval &a, const Interval &b)
{
    if(a.start==b.start)return a.end<b.end;
    return a.start<b.start;
}
class Solution {
public:
vector<Interval> merge(vector<Interval> &intervals) 
{    
    vector<Interval> res;
    if(intervals.empty())return res;    
    sort(intervals.begin(),intervals.end(),comp);
    Interval temp(intervals[0].start, intervals[0].end);
    for (int i=1;i<intervals.size();i++)
    {
        if (intervals[i].start<=temp.end)
        {
            temp.end=temp.end>intervals[i].end?temp.end:intervals[i].end;
        } 
        else
        {
            res.push_back(temp);
            temp.start=intervals[i].start;
            temp.end=intervals[i].end;
        }
    }
    res.push_back(temp);
    return res;
}
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) 
{
    vector<Interval> res;
    intervals.push_back(newInterval);
    res=merge(intervals);
    return res;
}
};
*/