*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;
        }
    }
    
}