lkeetcode 57. Insert Interval
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].
class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
if (intervals.empty())
{
intervals.push_back(newInterval);
return intervals;
}
if (newInterval.end < intervals[0].start)
{
intervals.insert(intervals.begin(), newInterval);
return intervals;
}
if (newInterval.end == intervals[0].start)
{
intervals[0].start= newInterval.start;
return intervals;
}
int k = 0;
while (k <intervals.size())
{
while (k <intervals.size() && intervals[k].end < newInterval.start)
k++;
if (k == intervals.size())
{
intervals.push_back(newInterval);
return intervals;
}
if (intervals[k].start > newInterval.end)
{
intervals.insert(intervals.begin() + k, newInterval);
return intervals;
}
intervals[k].start = newInterval.start < intervals[k].start ?
newInterval.start: intervals[k].start;
if (intervals[k].end >= newInterval.end)
return intervals;
int end = newInterval.end;
int kk = k;
while (kk < intervals.size() && end >= intervals[kk].end)
kk++;
if (kk == intervals.size())
{
intervals[k].end = end;
intervals.erase(intervals.begin() + k + 1, intervals.end());
return intervals;
}
if (intervals[kk].start <= end)
{
intervals[k].end = intervals[kk].end;
intervals.erase(intervals.begin() + k + 1, intervals.begin() + kk+1);
return intervals;
}
intervals[k].end = end;
intervals.erase(intervals.begin() + k + 1, intervals.begin() + kk);
return intervals;
}
}
};