Question
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]
.
Solution
Result: Accepted Time: 16 ms
Here should be some explanations.
class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
auto it = lower_bound(intervals.begin(), intervals.end(), newInterval,
[](const Interval& a, const Interval &b ){return a.start < b.start;});
vector<Interval> ret( intervals.begin(),it);
if(ret.size()&&ret.back().end >= newInterval.start)
ret.back().end = max(newInterval.end,ret.back().end);
else
ret.push_back(newInterval);
while(it!=intervals.end())
{
auto & tmp = ret.back();
if(tmp.end >= it->start)
tmp.end = max(tmp.end,it->end);
else
ret.push_back(*it);
++it;
}
return std::move(ret);
}
};
Complexity Analytics
- Time Complexity:
- Space Complexity: