问设置游程长度编码长度的最小限制

EN

1  def encode (plainText):
2      res=''
3      a=''
4      for i in plainText:
5          if a.count(i)>0:
6             a+=i
7          else:
8              if len(a)>3:
9                  res+="/" + str(len(a)) + a[0][:1]
10             else:
11                 res+=a
12                 a=i
13     return(res)

def encode (plainText):
    ## Lazy solution for some of the edge cases
    if plainText is None or len(plainText) == 0:
        return plainText

    ## We can join this together
    ## and have faster arbitrary 
    ## str length addition
    res=[]
    ## We only care about the last
    ## character, no need to save all
    prev_char=''
    ## And count it ourselves, its
    ## faster then calling count
    count=0
    for i in plainText:
        if i == prev_char:
            ## If its the same char
            ## increment count
            count += 1
            ## and then continue with next
            ## cycle. Avoid unneccasary indent.
            continue

        ## else
        if count > 2 or prev_char == '/':
            ## If more then 2 occurances
            ## we build the encoding.
            ## for 3 occurances the length is the same.
            ## '/' is a special character so we
            ## always encode it
            res.append(
                f"/{count}{prev_char}"
            )
        else:
            ## Otherwise just append the symbols
            res.append(prev_char*count)
        ## We observed at least 1 instance of i 
        count = 1
        ## Store for next comparison
        prev_char = i

    ## Now deal with last character.
    ## Without this your string would miss it.
    if count > 2 or prev_char == '/':
        res.append(
            f"/{count}{prev_char}"
        )
    else:
        res.append(prev_char*count)

    ## And build our string
    return ''.join(res)