Insert Interval
在已经排好序的区间中,插入一个新的区间,与merge的做法类似
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> res;
int size = intervals.size();
int i = ;
for (i = ; i < size; i++) {
if(newInterval.end < intervals[i].start){
break;
}else if(newInterval.start > intervals[i].end){
res.push_back(intervals[i]);
continue;
}else{
newInterval.start = min(newInterval.start,intervals[i].start);
newInterval.end = max(newInterval.end,intervals[i].end);
}
}
res.push_back(newInterval);
for(int j = i;j<size;j++){
res.push_back(intervals[j]);
}
return res;
}
};
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