[LeetCode][Java] Insert Interval

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题目：

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：

给定一组不互相覆盖的间隔数，将一个新的间隔数插入到他们之中（如果必要的话进行合并）.

你可以假设这些间隔数起初是根据他们的起始数排好序的.

样例1：

给定[1,3],[6,9]，插入并合并[2,5] 得到[1,5],[6,9].

样例2：

给定[1,2],[3,5],[6,7],[8,10],[12,16],插入并合并[4,9] 得到 [1,2],[3,10],[12,16].

这是因为新的间隔数[4,9] 覆盖了[3,5],[6,7],[8,10].

算法分析：

* 结合方法《Merge Intervals》
* 新加入 newInterval ,重新排序后然后按照上面的方法就好

AC代码：

<span style="font-size:12px;">public class Solution {    public List<Interval> insert(List<Interval> intervals, Interval newInterval)    {    intervals.add(newInterval);    if (intervals == null || intervals.size() <= 1)            return intervals;    return  merge(intervals);    }  public static List<Interval> merge(List<Interval> intervals)    {        if (intervals == null || intervals.size() <= 1)            return intervals;        // sort intervals by using self-defined Comparator        Collections.sort(intervals, new IntervalComparator());         ArrayList<Interval> result = new ArrayList<Interval>();         Interval prev = intervals.get(0);        for (int i = 1; i < intervals.size(); i++)         {            Interval curr = intervals.get(i);             if (prev.end >= curr.start)             {                // merged case            Interval merged = new Interval(prev.start, Math.max(prev.end, curr.end));            prev = merged;        }         else         {            result.add(prev);            prev = curr;        }        }        result.add(prev);        return result;   }}class IntervalComparator implements Comparator<Interval>{    public int compare(Interval i1, Interval i2)    {        return i1.start - i2.start;    }}</span>

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