链接
https://leetcode.cn/problems/license-key-formatting/description/
思路
字符串处理，没啥好说的...
代码
class Solution:
def licenseKeyFormatting(self, s: str, k: int) -> str:
valid_len, valid_s = self.get_length(s)
first_len = valid_len % k if valid_len % k != 0 else k
res = valid_s[:first_len]
for i in range(first_len, valid_len, k):
res += '-' + valid_s[i: i+k]
return res
def get_length(self, s):
res = 0
res_s = ''
for i in s:
if i != '-':
res += 1
res_s += i.upper()
return res, res_s
标签：格式化,res,self,len,valid,密钥,return,482,first From： https://www.cnblogs.com/bjfu-vth/p/17681493.html