将数字转换为Excel列标题

Question

这是Leetcode上的问题，被归类为“简单”。我已经去了几个小时，甚至打电话给同事。我无法弄清楚逻辑中的错。我不是在寻找解决问题的完全不同的解决方案。如果有人可以指出我的方法出了什么问题，我会很感激。

这个想法是将int转换为表示为Excel列标题的字符串（1 ='a'，2 ='b'... 27 ='aa'等）。这是我的代码，带有评论。该代码适用于许多输入（例如735 - ＆gt;'abg'），但对其他输入失败（例如，702 - ＆gt;'zz'）。

def numToCol(n):

  # Generate a key such that {1:'A', 2:'B', 3:'C'... 26:'Z'} (this works fine)
  key = {}
  for i in range(65, 91):
    key[i-64] = chr(i)

  # According to Wikipedia, the number of digits in the bijective base-k 
  # numeral representing a nonnegative integer n is floor(logk((n+1)*(k-1)))

  # exp = num of letters in final string
  exp = int(math.log((n+1)*25, 26)) # int() rounds it down
  col_name = ''
  num = n

  # The final number is represented by a(26**0) + b(26**1) + c(26**2) + ...
  # If exp = 3, then there are 3 letters in the final string, so we need to find
  # a(26**2) + b(26**1) + c(26**0).

  # If exp = 3, i iterates over 2, 1, 0.
  for i in range(exp-1, -1, -1):
    # factor = how many 26**i's there are in num, rounded down
    factor = int(num/(26**i))
    # add to the string the letter associated with that factor
    col_name = col_name + key[factor]
    # update the number for next iteration
    num = num - (factor*(26**i))

  return col_name

这是我写的一个函数，以转向反向方向（将字符串转换为int）。这有助于查看预期的结果应该是什么。它已确认工作。

def colToNum(string):
  '''
  Converts an upper-case string (e.g., 'ABC') to numbers as if they were
  column headers in Excel.
  '''
  key = {}
  for i in range(65, 91):
    key[chr(i)] = i-64

  new = []
  for idx, val in enumerate(string[::-1]):
    new.append(key[val] * 26**idx)

  return sum(new)

Answer 1

能够找到故障，但找不到解决问题的优雅方法。您可能已经注意到，每当您到达Z时，代码总是会失败，并且有多个字母。那是因为应有的因素，例如2和26，含义为Bz，变为3和0。这零导致了问题。这是设法使代码工作的黑客。除了if-statement（和num = num-（因子*（26 ** i））是否已移动），

for i in range(exp-1, -1, -1):
    # factor = how many 26**i's there are in num, rounded down
    
    factor = int(num/(26**i))
    num = num - (factor*(26**i))
    
    if num == 0 and i!=0:
        factor -=1
        num = 26

    # add to the string the letter associated with that factor
    col_name = col_name + key[factor]

这 一切都是相同的每当结果中有多个字母时，就会发生。

Answer 2

您是否检查这个答案？看起来像您要的地区。我也正在分享以下代码。

def excel_column_number_to_name(column_number):
    output = ""
    index = column_number - 1
    while index >= 0:
        character = chr((index % 26) + ord('A'))
        output = output + character
        index = index // 26 - 1

    return output[::-1]


for i in xrange(1, 1024):
    print "%4d : %s" % (i, excel_column_number_to_name(i))

Answer 3

减去1并生成剩余的模量26。0-25代表AZ。重复：

def num_to_col(num):
    if num < 1:
        raise ValueError('num must be >= 1')
    result = ''
    while num:
        num, rem = divmod(num - 1, 26)
        result = chr(ord('A') + rem) + result
    return result

# tests
for num in (1, 26, 26+1, 26*2, 26*2+1, 26*26+25,
            26*26+26, 26*26+26+1, 735,
            26*26*26+26*26+26, 26*26*26+26*26+26+1):
    print(f'{num:5} {num_to_col(num)}')

输出：

    1 A
   26 Z
   27 AA
   52 AZ
   53 BA
  701 ZY
  702 ZZ
  703 AAA
  735 ABG
18278 ZZZ
18279 AAAA