*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]

.

与上一题基本一样的解法。

public class Solution {
public List<Interval> insert(List<Interval> intervals, Interval newInterval)
{
List<Interval> res = new ArrayList<Interval>();
intervals.add(newInterval);
res = merge(intervals);
return res;
}

public List<Interval> merge(List<Interval> intervals)
{
if (intervals == null || intervals.size() <= 1) {
return intervals;
}
Collections.sort(intervals, new IntervalComparator());

List<Interval> res = new ArrayList<Interval>();
int i=0;
while(i<intervals.size())
{
Interval newint = new Interval();
int newstart = intervals.get(i).start;
int newend = intervals.get(i).end;
int j=i+1;
while(j<intervals.size()&&newend>=intervals.get(j).start)
{
newend = Math.max(newend,intervals.get(j).end);
j++;
}
newint.start=newstart;
newint.end=newend;
res.add(newint);
if(j!=i+1) i=j;
else i++;
}
return res;
}

private class IntervalComparator implements Comparator<Interval>
{
public int compare(Interval a, Interval b)
{
return a.start - b.start;
}
}

}