Insert Interval
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) {}
* };
*/
class Solution {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval)
{
vector<Interval> v;
int l=-1;
while(l+1<intervals.size() && intervals[l+1].end<newInterval.start) l=l+1;
int r=intervals.size();
while(r-1>=0 && intervals[r-1].start>newInterval.end) r=r-1;
for(int i=0;i<=l;i++)
v.push_back(intervals[i]);
//
l++;r--;
if(l<intervals.size() && intervals[l].start<newInterval.start) newInterval.start=intervals[l].start;
if(r>=0 && intervals[r].end>newInterval.end) newInterval.end=intervals[r].end;
v.push_back(newInterval);
//
r++;
for(int i=r;i<intervals.size();i++)
v.push_back(intervals[i]);
return v;
}
};
* 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:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval)
{
vector<Interval> v;
int l=-1;
while(l+1<intervals.size() && intervals[l+1].end<newInterval.start) l=l+1;
int r=intervals.size();
while(r-1>=0 && intervals[r-1].start>newInterval.end) r=r-1;
for(int i=0;i<=l;i++)
v.push_back(intervals[i]);
//
l++;r--;
if(l<intervals.size() && intervals[l].start<newInterval.start) newInterval.start=intervals[l].start;
if(r>=0 && intervals[r].end>newInterval.end) newInterval.end=intervals[r].end;
v.push_back(newInterval);
//
r++;
for(int i=r;i<intervals.size();i++)
v.push_back(intervals[i]);
return v;
}
};