如何在这个问题中迭代。实际上我发现从 A 到 Z 我们会收到索引 1-26，从 AA 到 ZZ 是 26+26^2。但是如何进行？

Question

In spreadsheets, columns are labeled as follows:

Label           Number
A               1
...             ...
Z               26
AA              27
...             ...
AZ              52
BA              53
...             ...
ZZ              702
AAA             703
...             ...
AAZ             728
ABA             729

A is the first column, B is the second column, Z is the 26th column, and so on. The three dots represent the missing labels and their column numbers. Using the table given above, deduce the mapping between column labels and their corresponding numbers. Accept the column label as input and print the corresponding column number as output.

I started as:

   n=input()
   alpha='ABCDEFGHIJKLMNOPQRSTUVWXYZ'


   if len(n)==1:
       print(alpha.index(n.upper())+1)
   elif len(n)==2

but I know it won't work.

Answer 1

First, we want to write a function which gives the position of a letter (which you already did more or less).

alpha='ABCDEFGHIJKLMNOPQRSTUVWXYZ'

def position(letter):
    return alpha.index(letter.upper())+1

You then have 2 cases:

• either the column input has one letter, so you just return the position of the letter in the alphabet
• either the input has 2 letters, in which case the column number is position(letter1) * 26 + position(letter2)

This gives us the following program:

def column_number(label):
    if len(label) == 1:
        return position(label)
    elif len(label) == 2:
        return 26*position(label[0]) + position(label[1])
    else:
        print("incorrect label format")

Answer 2

Since we have to continue as many times as the input requires, it's ideal to make use of recursion:

alpha = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

# retrieve position of a letter in the alphabet set
def position(letter):
  pos = alpha.index(letter) + 1
  return pos

# function to recursively find the column number  
def column_number(label, n):
  if n==1:
    return position(label)

  else:
    return ((26**(n-1)) * position(label[0])) + column_number(label[1:], n-1)

# main code
label = input().upper()
n = len(label)

col_n = column_number(label, n)
print(col_n)