你好,stackoverflow社区! 几周来,我一直在努力寻找一种方法来完成这项工作,而不寻求帮助(有点像个人的挑战),但我做不到,而且大学项目占用了我大部分的时间,不能做到这一点让我很沮丧,因为这看起来不太难,但我想不出怎么做。成为开发人员的第一课是学会与他人合作,对吗?所以我是来寻求帮助的。我有这个密码:
l = [None]*8
k = ['m','i','k','e']
for a in k:
for b in k:
for c in k:
l[0] = a
l[1] = b
l[2] = c
print(l)
输出如下:
['m', 'm', 'm', None, None, None, None, None]
['m', 'm', 'i', None, None, None, None, None]
['m', 'm', 'k', None, None, None, None, None]
['m', 'm', 'e', None, None, None, None, None]
['m', 'i', 'm', None, None, None, None, None]
['m', 'i', 'i', None, None, None, None, None]
['m', 'i', 'k', None, None, None, None, None]
['m', 'i', 'e', None, None, None, None, None]
['m', 'k', 'm', None, None, None, None, None]
['m', 'k', 'i', None, None, None, None, None]
['m', 'k', 'k', None, None, None, None, None]
['m', 'k', 'e', None, None, None, None, None]
['m', 'e', 'm', None, None, None, None, None]
['m', 'e', 'i', None, None, None, None, None]
['m', 'e', 'k', None, None, None, None, None]
['m', 'e', 'e', None, None, None, None, None]
['i', 'm', 'm', None, None, None, None, None]
['i', 'm', 'i', None, None, None, None, None]
['i', 'm', 'k', None, None, None, None, None]
['i', 'm', 'e', None, None, None, None, None]
['i', 'i', 'm', None, None, None, None, None]
['i', 'i', 'i', None, None, None, None, None]
['i', 'i', 'k', None, None, None, None, None]
['i', 'i', 'e', None, None, None, None, None]
['i', 'k', 'm', None, None, None, None, None]
['i', 'k', 'i', None, None, None, None, None]
['i', 'k', 'k', None, None, None, None, None]
['i', 'k', 'e', None, None, None, None, None]
['i', 'e', 'm', None, None, None, None, None]
['i', 'e', 'i', None, None, None, None, None]
['i', 'e', 'k', None, None, None, None, None]
['i', 'e', 'e', None, None, None, None, None]
['k', 'm', 'm', None, None, None, None, None]
['k', 'm', 'i', None, None, None, None, None]
['k', 'm', 'k', None, None, None, None, None]
['k', 'm', 'e', None, None, None, None, None]
['k', 'i', 'm', None, None, None, None, None]
['k', 'i', 'i', None, None, None, None, None]
['k', 'i', 'k', None, None, None, None, None]
['k', 'k', 'm', None, None, None, None, None]
['k', 'k', 'i', None, None, None, None, None]
['k', 'k', 'k', None, None, None, None, None]
['k', 'k', 'e', None, None, None, None, None]
['k', 'e', 'm', None, None, None, None, None]
['k', 'e', 'i', None, None, None, None, None]
['k', 'e', 'k', None, None, None, None, None]
['k', 'e', 'e', None, None, None, None, None]
['e', 'm', 'm', None, None, None, None, None]
['e', 'm', 'i', None, None, None, None, None]
['e', 'm', 'k', None, None, None, None, None]
['e', 'm', 'e', None, None, None, None, None]
['e', 'i', 'm', None, None, None, None, None]
['e', 'i', 'i', None, None, None, None, None]
['e', 'i', 'k', None, None, None, None, None]
['e', 'i', 'e', None, None, None, None, None]
['e', 'k', 'm', None, None, None, None, None]
['e', 'k', 'i', None, None, None, None, None]
['e', 'k', 'k', None, None, None, None, None]
['e', 'k', 'e', None, None, None, None, None]
['e', 'e', 'm', None, None, None, None, None]
['e', 'e', 'i', None, None, None, None, None]
['e', 'e', 'k', None, None, None, None, None]
['e', 'e', 'e', None, None, None, None, None]
我想做的是,有一个函数,它可以计算出nonelistI中有多少个空格想要。就像传递一个参数,选择要替换多少个无。我想象这个函数是如何工作的,在我的脑海中它有3个参数(无列表的大小,要替换的字母列表,要替换的无的数目)。我对递归很熟悉,但想不出用它实现这个的方法。提前感谢:)
递归是一种方法:
也可以使用标准工具:
顺便说一句,
k='mike'
的工作原理是一样的。你知道吗解决方法是使用递归。你知道吗
我将向您展示伪代码:
假设要填充到nonelist的索引6(或
l
)然后您只需调用
fillNoneList(l, k, 0, 6)
,就可以完成:)你传递数字和一个空的列表和列表k
相关问题 更多 >
编程相关推荐