LeetCode 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.

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题目有点长，其实很简单。题意是一个被dash插乱的字符串，现在要格式化它，规则就是以长度K分组，每组之间用dash连接，但是第一组的长度可以不是K，另外还要把小写字母转换为大写字母。 解法：首先把所有dash删掉，并把小写转换为大写，得到一个新的字符串strNew，然后计算strNew.size()%K，如果等于0，表明第一组的长度也为K；否则第一组的长度就是strNew.size()%K。接下来就是把每K长度的字符串用dash拼接起来了。完整代码如下： [cpp] class Solution { public: string licenseKeyFormatting(string S, int K) { string strNew = ""; for (int i = 0; i < S.size(); ++i) { if (S[i] == ‘-‘)continue; strNew += toupper(S[i]); } string ans = ""; int offset = strNew.size() % K; if (offset != 0) { ans = strNew.substr(0, offset) + "-"; } while (offset < strNew.size()) { ans += strNew.substr(offset, K) + "-"; offset += K; } return ans.substr(0, ans.size() – 1); } }; [/cpp] 本代码提交AC，用时9MS。]]>