我的代码没有类型错误。这来自Geek中用于极客的字符串模式。

Question

代码应按递减顺序打印。

示例：输入a= "Geek“

输出Geek Gee Ge G端

这实际上是极客给极客的，我正试图用不同的变体来解决这个问题。

#User function Template for python3
class Solution:
    def pattern(self, S):
        n = len(S)
        for i in range (0, n):
            for j in range(0, n - i) :
                print(S[j], end = "")
            print("")



#{ 
 # Driver Code Starts
#Initial Template for Python 3

if __name__ == '__main__':
    T=int(input())
    for i in range(T):
        S = input()
# ob = Solution()
        answer = ob.pattern(S)
        for value in answer:
            print(value)
            

# } Driver Code Ends

Traceback (most recent call last):
  File "/home/ba2f900c4eca91e4a091a2c7bf208eb5.py", line 22, in <module>
    for value in answer:
TypeError: 'NoneType' object is not iterable

Answer 1

使用两个循环通常不是最有效的，我建议切片。

如下所示：

class Solution:
    def pattern(self,a):
        return [a[i:] for i in range(len(a))]


# {
# Driver Code Starts
# Initial Template for Python 3

if __name__ == '__main__':
    a = input()
    ob = Solution()
    answer = ob.pattern(a)
    print(answer)

# } Driver Code Ends

Answer 2

您只需通过调用answer = ob.Pattern(S)打印一次结果，然后循环遍历answer，然后在此时为None。