License Key Formatting

题目：
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:
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.

思路：
这是个字符串处理题目，考察对字符串的操作，对字符串S，开始可去除“-”,然后用字符串长度除以K确定第一个“-”插入的位置。这里的细节就是要判断字符串长度len <= K的情况和 len > K的情况。

代码：

#include
#include
#include
using namespace std;
string keyFormatting(string s,int k){
    int i,len;
    //先删除"-"  
    while((i = s.find("-",0)) != string::npos){
        s.erase(i,1);
    }
    len = s.length();
    string newStr = "";
    if(len <= k){
        for(i = 0; i < s.length(); i++){
            char c = s.at(i);
            if(islower(c)){
                newStr += c-32;
            }
            newStr += c;
        }
        return newStr;
    }
    int num = len % k;
    for(i = 0; i < s.length(); i++){
        char c = s.at(i);
        if(islower(c)){
            c -= 32;
        }
        newStr += c;
        if(((i+1 == num) || ((i+1)%k == num))&& i != s.length() -1)
            newStr += "-";
    }
    return newStr;
}
int main(){
    int k;
    string s;
    cin>>k;
    cin>>s;
    cout<return 0;
} 

另一种解法：
从后面向前面扫，只在碰见“-”时才进行操作，每碰到K个非“-”字符就在newStr后添加一个“-”，再判断字符串若最后存在“-”否，存在就就删除之，最后将字符串逆序，注意逆序使用的方式，代码如下：

string format(string S,int K){
    int i,count;
    count = 0;
    string newStr = "";
    for(i= S.length()-1; i >=0; i--){
        if(S.at(i) != '-'){
            count++;
            if(islower(S.at(i)))
                S.at(i) -= 32; 
            newStr += S.at(i);
            //这一步可能导致最后还存有一个"-" 
            if(count % K == 0){
                newStr += "-";
            }
        }
    }
    //对字符串的某个位置操作时一定要注意是否可能越界 
    if(newStr.length() > 0 && newStr.at(newStr.length()-1) == '-')
        newStr.erase(newStr.length()-1,1);
    //字符串逆转函数,写法：reverse(str.begin(),str.end()),str。头文件为:algorithm 
    return reverse(newStr.begin(),newStr.end()),newStr;
}

PS：
判断字符c是否是小写字母的函数为islower(c)，需引入头文件为ctype.h