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题目及算法描述
Time complexity : O(nlgn).

code

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

class Solution:
    def merge(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: List[Interval]
        """
        intervals.sort(key = lambda interval:interval.start)
        merged_intervals = []
        
        for interval in intervals :
            if not merged_intervals or merged_intervals[-1].end < interval.start:
                merged_intervals.append(interval)
            else:
                merged_intervals[-1].end= max(merged_intervals[-1].end, interval.end)
        
        return merged_intervals

Details

list.sort()

list.sort() is a built-in function, sorting the list in place.
sort(*, key=None, reverse=False)

定义命名的关键字参数在没有可变参数的情况下不要忘了写分隔符*，否则定义的将是位置参数。

key specifies a function of one argument that is used to extract a comparison key from each list element .

i.e.: 不是用list中的元素来排序,而是各自派个代表出来火拼, 这个代表怎么选拔呢? 就按照你指定给key的函数名, (python 是函数式编程, 类似functor or function pointer)

list.sort() is stable.

this is helpful for sorting in multiple passes (for example, sort by department, then by salary grade).

sorted

list.sort() has no return, if you want to get a new list, please use built-in function sorted()
sorted(iterable, *, key=None, reverse=False)
Return a new sorted list from the items in iterable.
sorted() is also stable.