482. License Key Formatting 密钥格式化
作者： 负雪明烛 id： fuxuemingzhu 个人博客： http://fuxuemingzhu.cn/
You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.
Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase.
Given a non-empty string S and a number K, format the string according to the rules described above.
Input: S = "5F3Z-2e-9-w", K = 4 Output: "5F3Z-2E9W" Explanation: The string S has been split into two parts, each part has 4 characters. Note that the two extra dashes are not needed and can be removed.
Input: S = "2-5g-3-J", K = 2 Output: "2-5G-3J" Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.
- The length of string S will not exceed 12,000, and K is a positive integer.
- String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
- String S is non-empty.
class Solution(object): def licenseKeyFormatting(self, S, K): """ :type S: str :type K: int :rtype: str """ S = S.upper() groups = ''.join(S.split('-')) bias = len(groups) % K devides = len(groups) / K answer = groups[:bias] answer += '-' if bias != 0 else '' for i in range(devides): answer += groups[i*K+bias : (i+1)*K+bias] + '-' return answer[:-1]
class Solution(object): def licenseKeyFormatting(self, S, K): """ :type S: str :type K: int :rtype: str """ res =  s = "".join(S.split("-")).upper() N = len(s) if N % K != 0: res.append(s[: N % K]) for i in range(N % K, N, K): res.append(s[i : i + K]) return "-".join(res)
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