标签：can 提交 bee between form tin ++ 增加 move
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:
string licenseKeyFormatting(string S, int K)
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。
leetcode-482-License Key Formatting
标签：can 提交 bee between form tin ++ 增加 move
原文地址：https://www.cnblogs.com/king-3/p/9159194.html
