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].

把一个区间插入一个已排序的无重叠的区间， 如果重叠的话，合并之

方法一：有点类似于快排的方法

1 class Solution(object):2     def insert(self, intervals, newInterval):3         s,e = newInterval.start,newInterval.end4         small = [ i for i in intervals if i.ende ]6         if small + large != intervals:7             s = min(s,intervals[len(small)].start)8             e = max(e,intervals[~len(large)].end)9         return small+[Interval(s,e)]+large

方法二：和上面方法一样，不过这个代码太帅了！！！只需要遍历一次

1 class Solution(object): 2     def insert(self, intervals, newInterval): 3         s,e = newInterval.start,newInterval.end 4         parts = mid,left,right=[],[],[] 5         for v in intervals: 6             parts[(v.end
     
      e)].append(v) 7         if mid: 8             s = min(s,mid[0].start) 9             e = max(e,mid[-1].end)10         return left+[Interval(s,e)]+right