[LeetCode]题解（python）：057-Insert Interval

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题目来源

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https://leetcode.com/problems/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].

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题意分析

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Input:a list of intervals and a new interval

Output:insert the new interval into the list and merge if necessary

Conditions:将新的interval插入进去，如果需要合并则合并

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题目思路

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感觉跟merge interval很像，所以直接append list中，然后排序按照上一题merge的方法，然后过了……

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AC代码（Python）

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 # Definition for an interval.
 # class Interval(object):
 #     def __init__(self, s=0, e=0):
 #         self.start = s
 #         self.end = e

 class Solution(object):
     def insert(self, intervals, newInterval):
         """
         :type intervals: List[Interval]
         :type newInterval: Interval
         :rtype: List[Interval]
         """
         intervals.append(newInterval)
         intervals.sort(key = lambda x:x.start)

         length = len(intervals)
         res = []

         if length == 0:
             res.append(newInterval)
             return res

         res.append(intervals[0])
         for i in range(1,length):
             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