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:

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.

 

要完成的函数：

string licenseKeyFormatting(string S, int K) 

 

说明：

1、给定一个字符串S和一个正数K，字符串中只含有字母、数字和破折号，比如5F3Z-2e-9-w。

要求将字符串的格式重新编排，使得被破折号隔开的每个部分含有K个字符，除了最开始的第一个部分可以是小于等于K个字符的，但必须至少有一个字符。

同时要求字符串中的小写字母转化为大写字母。

比如字符串是5F3Z-2e-9-w，K=4，那么重新编排完格式是5F3Z-2E9W。

如果字符串是9-5F3Z-2e-9-w，K=4，那么重新编排完格式就是9-5F3Z-2E9W。

 

2、题意清晰，这道题我们要从字符串的最后开始处理，这样比较方便。

代码如下：（附详解）

string licenseKeyFormatting(string S, int K)     {        string res;//最后要返回的字符串        int i=S.size()-1,count=0;        while(i>=0)        {            if(S[i]!='-')//如果这个字符是字母或者数字            {                count++;                if(count<=K)//如果还没有达到K个                {                    res+=char(toupper(S[i]));//大小写转换，增加到res中                    i--;                }                else//如果达到了K个                {                    res+='-';//插入'-'                    count=0;//重新开始计数                }            }            else//如果这个字符是'-'                i--;        }        reverse(res.begin(),res.end());//最后反转一下，就是我们要的字符串。        return res;    }

上述代码实测12ms，beats 98.58% of cpp submissions。

 

3、一些其他说明：

可能有的同学写的也是跟笔者一样的2中的代码，只不过在res+=char(toupper(S[i]))这里，改成了res=char(toupper(S[i]))+res。这样最后就不用reverse了。

但这样提交了之后会发现花费时间巨大，是2中代码花费时间的20倍左右。

原因是res=char(toupper(S[i]))+res处理的时候，可以看做是重新定义一个临时字符串，把S[I]的值放进去，然后再增加了res，最后把临时字符串赋给了res。

而res+=char(toupper(S[i]))是直接在res后面增加了S[i]，类似于vector.push_back()。

两者相比较起来，还是2中的方法快速，最后做一下reverse其实花费时间也不多。