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]

.

Runtime: 588ms

/**
* 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> result;

if(intervals.empty()){
result.push_back(newInterval);
return result;
}

int i = 0;
int n = intervals.size();
for(; i < n; i++){
if(newInterval.end < intervals[i].start){
//result.push_back(newInterval);
break;
}
else if(newInterval.start > intervals[i].end){
result.push_back(intervals[i]);
continue;
}
else{
newInterval.start = min(newInterval.start, intervals[i].start);
newInterval.end = max(newInterval.end, intervals[i].end);
}
}

result.push_back(newInterval);

for(; i < intervals.size(); i++)
result.push_back(intervals[i]);

return result;
}
};