Desicription
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].
Solution
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class {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
vector<Interval> res;
auto it = intervals.begin();
for(; it != intervals.end(); it++){
if(it->start > newInterval.end)
break;
else if(it->end < newInterval.start)
res.push_back(*it);
else
newInterval.start = min(newInterval.start, it->start), newInterval.end = max(newInterval.end, it->end);
}
res.push_back(newInterval);
for(; it != intervals.end(); it++)
res.push_back(*it);
return res;
}
};
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