# 482. License Key Formatting 密钥格式化

@TOC

## # 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 = 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.
``````

Example 2:

``````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.
``````

Note:

1. The length of string S will not exceed 12,000, and K is a positive integer.
2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
3. String S is non-empty.

## # 解题方法

``````class Solution(object):
"""
:type S: str
:type K: int
:rtype: str
"""
S = S.upper()
groups = ''.join(S.split('-'))
bias = len(groups) % K
devides = len(groups) / K
answer += '-' if bias != 0 else ''
for i in range(devides):
answer += groups[i*K+bias : (i+1)*K+bias] + '-'
``````

``````class Solution(object):
"""
: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)
``````

## # 日期

2018 年 2 月 1 日 2018 年 11 月 22 日 —— 感恩节快乐～