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) {}
* };
*/
struct Interval
{
int start;
int end;
Interval() : start(0), end(0) {}
Interval(int s, int e) : start(s), end(e) {}
};
bool operator <(const Interval a, const Interval b) { return a.start < b.start; }
class Solution {
public:
std::vector<Interval> insert(std::vector<Interval> &intervals, Interval newInterval) {
std::vector<Interval> res;
intervals.push_back(newInterval);
std::sort(intervals.begin(),intervals.end());
int left = intervals[0].start, right = intervals[0].end;
for (int i = 1; i < intervals.size(); i++)
{
if(intervals[i].start <= right)
{
right = std::max(right,intervals[i].end);
}
else
{
res.push_back(Interval(left,right));
left = intervals[i].start;
right = intervals[i].end;
}
}
res.push_back(Interval(left,right));
return res;
}
};