leetcode-482-License Key Formatting
題目描述：
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:
- The length of string S will not exceed 12,000, and K is a positive integer.
- String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
- String S is non-empty.
要完成的函數：
string licenseKeyFormatting(string S, int K)
說明：
1、給定一個字符串S和一個正數K，字符串中只含有字母、數字和破折號，比如5F3Z-2e-9-w。
要求將字符串的格式重新編排，使得被破折號隔開的每個部分含有K個字符，除了最開始的第一個部分可以是小於等於K個字符的，但必須至少有一個字符。
同時要求字符串中的小寫字母轉化為大寫字母。
比如字符串是5F3Z-2e-9-w，K=4，那麽重新編排完格式是5F3Z-2E9W。
如果字符串是9-5F3Z-2e-9-w，K=4，那麽重新編排完格式就是9-5F3Z-2E9W。
2、題意清晰，這道題我們要從字符串的最後開始處理，這樣比較方便。
代碼如下：（附詳解）
string licenseKeyFormatting(string S, int K)
{
string res;//最後要返回的字符串
int i=S.size()-1,count=0;
while(i>=0)
{
if(S[i]!=‘-‘)//如果這個字符是字母或者數字
{
count++;
if(count<=K)//如果還沒有達到K個
{
res+=char(toupper(S[i]));//大小寫轉換，增加到res中
i--;
}
else//如果達到了K個
{
res+=‘-‘;//插入‘-‘
count=0;//重新開始計數
}
}
else//如果這個字符是‘-‘
i--;
}
reverse(res.begin(),res.end());//最後反轉一下，就是我們要的字符串。
return res;
}
上述代碼實測12ms，beats 98.58% of cpp submissions。
3、一些其他說明：
可能有的同學寫的也是跟筆者一樣的2中的代碼，只不過在res+=char(toupper(S[i]))這裏，改成了res=char(toupper(S[i]))+res。這樣最後就不用reverse了。
但這樣提交了之後會發現花費時間巨大，是2中代碼花費時間的20倍左右。
原因是res=char(toupper(S[i]))+res處理的時候，可以看做是重新定義一個臨時字符串，把S[I]的值放進去，然後再增加了res，最後把臨時字符串賦給了res。
而res+=char(toupper(S[i]))是直接在res後面增加了S[i]，類似於vector.push_back()。
兩者相比較起來，還是2中的方法快速，最後做一下reverse其實花費時間也不多。
leetcode-482-License Key Formatting