方法有三：

1. 将列表构造为集合，再判断长度

2. 用一个元素与所有元素比较

3. 比较列表第一个元素的个数和列表长度

   Pythonic ways of checking if all

   items in a list are equal:

   lst = ['a', 'a', 'a']

   len(set(lst)) == 1 True

   all(x == lst[0] for x in lst) True

   lst.count(lst[0]) == len(lst) True

   I ordered those from "most Pythonic" to "least Pythonic"

   and "least efficient" to "most efficient".

   The len(set()) solution is idiomatic, but constructing

   a set is less efficient memory and speed-wise.

三个方法的排序，语言地道性由高到低，效率由低到高。构造一个集合就是用内存换速度的。