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].
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
int start = 0, end = 0;
for (; start < intervals.size() && intervals[start].end < newInterval.start; start++);
for (end = start; end < intervals.size() && intervals[end].start <= newInterval.end; end++);
end--;
if (start <= end && start < intervals.size())
{
intervals[start].start = min(intervals[start].start, newInterval.start);
intervals[start].end = max(intervals[end].end, newInterval.end);
if (start < end)
intervals.erase(intervals.begin() + start + 1, intervals.begin() + end + 1);
}
else
intervals.insert(intervals.begin() + start, newInterval);
return intervals;
}
本文介绍了一种算法,用于将新区间插入到一组不重叠区间中,并在必要时进行合并。算法假定初始区间已按开始时间排序。
1495

被折叠的 条评论
为什么被折叠?



