原题地址：https://oj.leetcode.com/problems/insert-interval/
题意：
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].
解题思路：最简单的方法是将要插入的区间和原来的区间合在一起排序，然后按照merge intervals的方法来编程。
代码：
# Definition for an interval. # class Interval: # def __init__(self, s=0, e=0): # self.start = s # self.end = e class Solution: # @param intervals, a list of Intervals # @param newInterval, a Interval # @return a list of Interval # @should rewrite without sort!!! def insert(self, intervals, newInterval): intervals.append(newInterval) intervals.sort(key = lambda x:x.start) length=len(intervals) res=[] for i in range(length): if res==[]: res.append(intervals[i]) else: size=len(res) if res[size-1].start<=intervals[i].start<=res[size-1].end: res[size-1].end=max(intervals[i].end, res[size-1].end) else: res.append(intervals[i]) return res
[leetcode]Insert Interval @ Python,布布扣,bubuko.com
[leetcode]Insert Interval @ Python
原文：http://www.cnblogs.com/zuoyuan/p/3782048.html