LeetCode 57. Insert Interval---Python实现

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]

.

这题与上题类似，直接将新的Interval添加到Interval中再次调用上题的方法就可以了.参考了南郭子綦的方法

# 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)    #x就是代表inverval里面的一个属性啊
        res=[]
        for i in range(len(intervals)):
            if res == []:
                res.append(intervals[i])
            else:
                size=len(res)
                if res[size-1].start<=intervals[i].start<=res[size-1].end: #size-1不能写成了i
                    res[size-1].end=max(res[size-1].end,intervals[i].end)
                else:
                    res.append(intervals[i])
        return res