LeetCode 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].

题意 ：给出一个有序区间段，在插入一个区间段后，求其合并后的有序区间段

思路：确切的说，有四种情况：

（1）原来为空

（2）插入的区间段小于原来的第一个区间段

（3）插入的区间段大于原来的最后一个区间段

（4）找到插入区间开始和结束位置在原来的有序区间段的索引位置，找开始位置时，如果没有找到，就以大于开始位置的索引为准。而在找结束位置时，如果没有找到， 就以小于结束位置的索引为准

代码如下：

class Solution
{
    public List<Interval> insert(List<Interval> intervals, Interval newInterval)
    {
        List<Interval> ans = new ArrayList<Interval>();
        int size = intervals.size();
        if (0 == size)
        {
            ans.add(newInterval);
            return ans;
        }

        if (newInterval.end < intervals.get(0).start)
        {
            ans.add(newInterval);
            ans.addAll(intervals);
            return ans;
        }
        else if (newInterval.start > intervals.get(size - 1).end)
        {
            ans.addAll(intervals);
            ans.add(newInterval);
            return ans;
        }

        int index1 = 0, index2 = size - 1;

        for (int i = 0; i < size; i++)
        {
            Interval cur = intervals.get(i);
            if (newInterval.start >= cur.start && newInterval.start <= cur.end)
            {
                index1 = i;
            }
            else if (i != size - 1)
            {
                Interval next = intervals.get(i + 1);
                if (newInterval.start > cur.end && newInterval.start < next.start)
                {
                    index1 = i + 1;
                }
            }

            if (newInterval.end >= cur.start && newInterval.end <= cur.end)
            {
                index2 = i;
            }
            else if (i != size - 1)
            {

                Interval next = intervals.get(i + 1);
                if (newInterval.end > cur.end && newInterval.end < next.start)
                {
                    index2 = i;
                }
            }
        }

        for (int i = 0; i < index1; i++)
        {
            Interval cur = intervals.get(i);
            ans.add(cur);
        }

        Interval tmp = new Interval();
        tmp.start = Math.min(intervals.get(index1).start, newInterval.start);
        tmp.end = Math.max(intervals.get(index2).end, newInterval.end);
        ans.add(tmp);

        for (int i = index2 + 1; i < size; i ++)
        {
            Interval cur = intervals.get(i);
            ans.add(cur);
        }
        return ans;
    }
}