题目描述：

LeetCode 482. License Key Formatting

Now you are given a string S, which represents a software license key which we would like to format. The string S is composed of alphanumerical characters and dashes. The dashes split the alphanumerical characters within the string into groups. (i.e. if there are M dashes, the string is split into M+1 groups). The dashes in the given string are possibly misplaced.

We want each group of characters to be of length K (except for possibly the first group, which could be shorter, but still must contain at least one character). To satisfy this requirement, we will reinsert dashes. Additionally, all the lower case letters in the string must be converted to upper case.

So, you are given a non-empty string S, representing a license key to format, and an integer K. And you need to return the license key formatted according to the description above.

Example 1:

Input: S = "2-4A0r7-4k", K = 4

Output: "24A0-R74K"

Explanation: The string S has been split into two parts, each part has 4 characters.

Example 2:

Input: S = "2-4A0r7-4k", K = 3

Output: "24-A0R-74K"

Explanation: The string S has been split into three parts, each part has 3 characters except the first part as it could be shorter as said 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.

题目大意：

给定字符串S表示一个软件的授权码。S由字母数字以及破折号组成。破折号将字母数字分成若干组。（如果有M个破折号，就分成M+1组）。破折号的位置可能不正确。

现在将S分成若干组长度为K的字符串（第一组的长度可以比K小，但是不能为0）。为满足这个条件，我们重新插入破折号。此外，小写字母应该转化为大写字母。

给定非空字符串S，表示待格式化的软件授权码，以及一个整数K。返回按照上述描述重新格式化的授权码。

注意：

1. S的长度不会超过12000，K是正整数
2. 字符串S只包含字母和数字（a-z，A-Z，0-9）以及破折号（-）
3. 字符串S是非空的

解题思路：

字符串模拟

将S的开头部分用'#'补足，使其长度是K的整数倍，返回时将#删去

Python代码：

class Solution(object):
    def licenseKeyFormatting(self, S, K):
        """
        :type S: str
        :type K: int
        :rtype: str
        """
        S = S.replace('-', '').upper()
        if len(S) % K:
            S = '#' * (K - len(S) % K) + S
        return '-'.join(S[x:x + K] for x in range(0, len(S), K)).replace('#', '')

 

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