一遍扫描数组,遇到新的有交叉的更新newInterval
没有交叉后就直接插入。
/**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<Interval> r;
int i=0;
for(;i<intervals.size();i++)
{
if(intervals[i].end<newInterval.start)
{
r.push_back(intervals[i]);
}
else if(intervals[i].start<=newInterval.end
&& intervals[i].end>=newInterval.start)
{
newInterval.start=min(intervals[i].start,newInterval.start);
newInterval.end=max(intervals[i].end,newInterval.end);
}
else
{
break;
}
}
r.push_back(newInterval);
for(;i<intervals.size();i++)
{
r.push_back(intervals[i]);
}
return r;
}
};