Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

/**
 * 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> merge(vector<Interval> &intervals) 
    {
        vector<Interval> res;
        vector<Interval>& in = intervals;
        int n = in.size();
        if(n == 0)
            return res;
        if(n == 1)
            return in;
        heap_sort(in);
        Interval temp = in[0];
        for(int i=1; i<n; i++)
        {
            if(temp.end < in[i].start)
            {
                res.push_back(temp);
                temp = in[i];
            }
            else
            {
                int start = temp.start<in[i].start? temp.start:in[i].start;
                int end = temp.end>in[i].end? temp.end:in[i].end;
                temp.start = start;
                temp.end = end;
            }
        }
        res.push_back(temp);
        return res;
    }
    void swap(vector<Interval>& in, int i, int j)
    {
        int s = in[i].start;
        int e = in[i].end;
        in[i].start = in[j].start;
        in[i].end = in[j].end;
        in[j].start = s;
        in[j].end = e;
    }
    void sink(vector<Interval>& in, int loc, int end)
    {
        while(loc<end)
        {
            int k =2*loc+1;
            if(k>end)
                return;
            if(k<end && in[k].start<in[k+1].start)
                k++;
            if(in[loc].start>in[k].start)
                return;
            swap(in,loc,k);
            loc = k;
        }
    }
    void heap_sort(vector<Interval>& in)
    {
        int n = in.size()-1;
        for(int i=(n-1)/2; i>=0; i--)
            sink(in,i,n);
        while(n>0)
        {
            swap(in,0,n);
            n--;
            sink(in,0,n);
        }
    }
};

LeetCode--Merge Intervals