将数字转换为 Excel 列 headers

Question

这是 LeetCode 上的一道题，被归类为“简单”。我已经为此工作了几个小时，甚至还请来了一位同事。我无法找出我逻辑中的错误。我不是在寻找完全不同的问题解决方案。如果有人能指出我的方法有什么问题，我将不胜感激。

想法是将 int 转换为表示为 Excel 列的字符串 header (1='A', 2='B' ... 27='AA'，等等）。这是我的代码，带有注释。该代码适用于许多输入（例如，735 -> 'ABG'），但对其他输入无效（例如，702 -> '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

这是我编写的一个反向函数（将字符串转换为 int）。这有助于了解预期结果应该是什么。确认有效。

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

你检查过this answer了吗？那个看起来像你所要求的。我也分享如下代码。

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 2

找到问题所在，但没有找到优雅的解决方法。您可能已经注意到，每当您到达 Z 并且有多个字母时，代码总是会失败。那是因为应该是2和26的因数，意思是BZ，变成了3和0。这个零导致了问题。这是设法使代码工作的技巧。除了 if-statement（并且 num = num - (factor*(26**i)) 已向上移动）

之外，其他一切都一样

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]

这个 if-statements 只是修复了当结果中有多个字母时发生的这些“Z”情况。

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