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]
.
比较tricky的就是先排序,然后比较头尾是不是需要merge,如需要比较尾部是不是需要,否则就不merge, push
/**
* 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:
static bool comp(Interval& l, Interval&r){
return l.start<r.start;
}
vector<Interval> merge(vector<Interval> &intervals) {
sort(intervals.begin(),intervals.end(),comp);
vector<Interval> res;
if (intervals.size()==0)
return res;
res.push_back(intervals[0]);
int end=0;
for (int i=1; i<intervals.size();i++){
if (intervals[i].start<=res[end].end){
if (intervals[i].end>=res[end].end)
res[end].end=intervals[i].end;
} else{
res.push_back(intervals[i]);
end++;
}
}
return res;
}
};