[Java]LeetCode57 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：该题与56题非常相似，只是题意给出每一个区间的start是递增的。所以我们不须要排序。该题须要我们加入一个区间。然后进行融合。

这题在56题的基础上添加了区间推断的复杂度。

包括例如以下如所看到的的六种情况。



而这六种情况又能够合并成三种解决方案。看例如以下代码：

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
       // 先推断newInterval是否在intervals的范围内
        if (newInterval == null)
            return intervals;
        int len = intervals.size();
        if (len == 0)
        {
            intervals.add(newInterval);
            return intervals;
        }
        List<Interval> res=new ArrayList<Interval>();
        for(Interval interval:intervals)
        {
            if(interval.end<newInterval.start)//newInterval在中间的情况
            {
                res.add(interval);
            }else if(interval.start>newInterval.end)//newInterval插入最前端的情况
            {
                res.add(newInterval);
                newInterval=interval;//这个地方非常重要。就是找到了待插入区间位置。指定新的newInterval，由于intervals中的区间也可能有相交的地方，须要融合。
            }else if(interval.start<=newInterval.end||interval.end>=newInterval.start)//有重合部分的四种情况
            {
                newInterval=new Interval(Math.min(interval.start,newInterval.start),Math.max(interval.end,newInterval.end));
            }
        }
        res.add(newInterval);
        return res;
    }
}