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) {}
* };
*/
bool cmp(const Interval &a, const Interval &b)
{
return a.start < b.start;
}
class Solution {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
intervals.push_back(newInterval);
return merge(intervals);
}
vector<Interval> merge(vector<Interval> &intervals) {
sort(intervals.begin(), intervals.end(), cmp);
vector<Interval> res;
if(intervals.empty()) return res;
vector<Interval>::iterator it = intervals.begin();
Interval temp = *it++;
if(it == intervals.end())
res.push_back(temp);
else
{
while(it != intervals.end())
{
while((*it).start <= temp.end && it != intervals.end())
{
temp.end = max(temp.end, (*it).end);
++it;
}
res.push_back(temp);
temp = *it;
}
}
return res;
}
};