【LeetCode】482. License Key Formatting 解题报告（Python）

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作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目描述

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.

Example 1:

Input: S = "5F3Z-2e-9-w", K = 4Output: "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.

Example 2:

Input: S = "2-5g-3-J", K = 2Output: "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.

Note:

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.

题目大意

现在有一些用-分割的字符串，需要重新安排，使得除了第一个之外，其他的-分割的字符串长度都是K。另外需要全部转成大写字符。

解题方法

注意，这个题的意思是右边的序列要都是K个的，最左边如果不够就不够了，剩多少写多少。

首先计算第一个应该占据了多少个字符，然后看后面的应该是等长的。并且和原来的-的划分情况是无关的。每个片的个数都要是K。
字符串切片结束的长度大于自身长度也可以。

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]

二刷的时候，写的Python代码如下：

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)