概述
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:
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>::iterator it = intervals.begin();
while (it != intervals.end()) {
if (newInterval.end < it->start) {
intervals.insert(it, newInterval);
return intervals;
} else if (newInterval.start > it->end) {
it++;
continue;
} else {
newInterval.start = min(newInterval.start, it->start);
newInterval.end = max(newInterval.end, it->end);
it = intervals.erase(it);
}
}
intervals.push_back(newInterval);
return intervals;
}
};
最后
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