内置功能最少，假设是面试设置

array = [1, 2, 3, 4, 5, 6,7, 8]
inverse = [] #create container for inverse array
length = len(array)  #to iterate later, returns 8 
counter = length - 1  #because the 8th element is on position 7 (as python starts from 0)

for i in range(length): 
   inverse.append(array[counter])
   counter -= 1
print(inverse)

这个类使用Python魔法方法和迭代器进行反转，并反转列表:

class Reverse(object):
    """ Builds a reverse method using magic methods """

    def __init__(self, data):
        self.data = data
        self.index = len(data)


    def __iter__(self):
        return self

    def __next__(self):
        if self.index == 0:
            raise StopIteration

        self.index = self.index - 1
        return self.data[self.index]


REV_INSTANCE = Reverse([0, 10, 20, 40])

iter(REV_INSTANCE)

rev_list = []
for i in REV_INSTANCE:
    rev_list.append(i)

print(rev_list)  

输出

[40, 20, 10, 0]

def reverse(text):
    output = []
    for i in range(len(text)-1, -1, -1):
        output.append(text[i])
    return output

>>> xs = [0, 10, 20, 40]
>>> xs[::-1]
[40, 20, 10, 0]

这里将解释扩展片语法。请参见文档。

有3种方法可以获得反向列表:

切片方法1:reversed_array = array[-1::-1] 切片方法二: Reversed_array2 = array[::-1] 使用内置函数:reversed_array = array.reverse()

第三个函数实际上颠倒了列表对象。这意味着不维护原始数据的副本。如果您不想维护旧版本，这是一个很好的方法。但如果你想要原始和反转的版本，这似乎不是一个解决方案。

The cute Solution


class List_reverse():
    def reverse_list(self):
        list = [1, 2, 3, 4, 5, 6]
        list1 = []
        for i in range(1, len(list)+1):
            list1.append(list[-i])
        print(list1)


if __name__ == "__main__":
    obj = List_reverse()
    obj.reverse_list()
    enter code here