Given a collection of intervals, merge all overlapping intervals.
Example
Given intervals => merged intervals:
[ [
[1, 3], [1, 6],
[2, 6], => [8, 10],
[8, 10], [15, 18]
[15, 18] ]
]
Challenge
O(n log n) time and O(1) extra space.
/**
* Definition of Interval:
* classs Interval {
* int start, end;
* Interval(int start, int end) {
* this->start = start;
* this->end = end;
* }
*/
class Solution {
static bool compFunc(const Interval &i1, const Interval& i2)
{
return i1.start < i2.start;
}
public:
vector<Interval> merge(vector<Interval> &intervals) {
int size = intervals.size();
vector<Interval> res;
if(0 == size) return res;
sort(intervals.begin(), intervals.end(), compFunc);
res.push_back(intervals[0]);
for(int i=1;i<size;i++)
{
if(!(res.back().start>intervals[i].end
|| res.back().end < intervals[i].start)){
res.back().start = min(res.back().start, intervals[i].start);
res.back().end = max(res.back().end, intervals[i].end);
}
else{
res.push_back(intervals[i]);
}
}
return res;
}
};