CS61A_未解决

 def close(n, smallest=10, d=10):
     """ A sequence is near increasing if each element but the last two is smaller than all elements
     following its subsequent element. That is, element i must be smaller than elements i + 2, i + 3, i + 4, etc.
     Implement close, which takes a non-negative integer n and returns the largest near increasing sequence
     of digits within n as an integer. The arguments smallest and d are part of the implementation; you must
     determine their purpose. The only values you may use are integers and booleans (True and False) (no lists, strings, etc.).
     Return the longest sequence of near-increasing digits in n.
     >>> close(123)
     123
     >>> close(153)
     153
     >>> close(1523)
     153
     >>> close(15123)
     1123
     >>> close(11111111)
     11
     >>> close(985357)
     557
     >>> close(14735476)
     143576
     >>> close(812348567)
     1234567
     """
     if n == 0:
       return ______
     no = close(n//10, smallest, d)
     if smallest > ______:
         yes = ______
         return ______(yes, no)
     return ___

def close(n, smallest=10, d=10):
    """ A sequence is near increasing if each element but the last two is smaller than all elements
    following its subsequent element. That is, element i must be smaller than elements i + 2, i + 3, i + 4, etc.
    Implement close, which takes a non-negative integer n and returns the largest near increasing sequence
    of digits within n as an integer. The arguments smallest and d are part of the implementation; you must
    determine their purpose. The only values you may use are integers and booleans (True and False) (no lists, strings, etc.).
    Return the longest sequence of near-increasing digits in n.
    >>> close(123)
    123
    >>> close(153)
    153
    >>> close(1523)
    153
    >>> close(15123)
    1123
    >>> close(11111111)
    11
    >>> close(985357)
    557
    >>> close(14735476)
    143576
    >>> close(812348567)
    1234567
    """
    if n == 0:
      return 0
    no = close(n//10, smallest, d)
    if smallest > n % 10:
        yes = 10 * close(n//10, d, min(d, n%10)) + n%10
        return max(yes, no)
    return no