题目描述:
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:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
Example 2:
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
由于区间都已经排好序,所以从第一个区间开始,判断是否和插入区间相交,直到找到相交的区间之后进行合并,将合并的新区间和下一个区间比较,直到不再相交为止。
/**
* 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;
int i=0;
while(i<intervals.size())
{
if(intervals[i].end<newInterval.start)
{
result.push_back(intervals[i]);
i++;
}
else break;
}
while(i<intervals.size()&&intersect(intervals[i],newInterval))
{
newInterval.start=min(newInterval.start,intervals[i].start);
newInterval.end=max(newInterval.end,intervals[i].end);
i++;
}
result.push_back(newInterval);
while(i<intervals.size())
{
result.push_back(intervals[i]);
i++;
}
return result;
}
bool intersect(Interval a, Interval b)
{
if(a.end<b.start||b.end<a.start) return false;
else return true;
}
};