Given
a collection of intervals, merge all overlapping intervals.
/**
* 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 {
private:
void quicksort(vector<Interval> & datas,
int left,
int right)
{
if(left>=right)
return;
int l=left;
int r=right;
int m=datas[(l+r)/
2].start;
while(l<=r)
{
while(datas[l].start<m) l++;
while(datas[r].start>m) r--;
if(l>right || r<left || l>=r)
break;
Interval tmp=datas[l];
datas[l]=datas[r];
datas[r]=tmp;
l++;r--;
}
quicksort(datas,left,l-
1);
quicksort(datas,r+
1,right);
}
inline
bool checkoverlap(
const Interval& interval1,
const Interval& interval2)
{
if(interval1.start<=interval2.start && interval1.end>=interval2.end)
return true;
if(interval1.start<=interval2.start && interval1.end>=interval2.start)
return true;
if(interval1.start<=interval2.end && interval1.end>=interval2.end)
return true;
return false;
}
inline
void mergeinterval(Interval& interval1,
const Interval& interval2)
{
if(interval1.start>interval2.start) interval1.start=interval2.start;
if(interval1.end<interval2.end) interval1.end=interval2.end;
}
public:
vector<Interval> merge(vector<Interval> &intervals)
{
vector<Interval> v;
if(intervals.size()==
0)
return v;
//sort by start
quicksort(intervals,
0,intervals.size()-
1);
//merge
int index=
0;
v.push_back(intervals[
0]);
for(
int i=
1;i<intervals.size();i++)
{
if(checkoverlap(v[index],intervals[i]))
{
mergeinterval(v[index],intervals[i]);
}
else {
v.push_back(intervals[i]);
index++;
}
}
return v;
}
};