第一版：

    int log = 0;
    string ele="";
    string re="";
    for (int i = S.length()-1; i>=0; i--) {
        if (S[i]!='-') {
            log++;
            S[i]=toupper(S[i]);
            ele = S[i]+ele;
        }
        if (log==K||i==0) {
            if (i==0) {
                re = ele+re;
            }else{
                re = '-'+ele+re;
            }
            log = 0;
            ele = "";
        }
    }
    return re[0]=='-'?re.substr(1,re.length()-1):re;

第二版：运用栈

class Solution {
public:
    string licenseKeyFormatting(string S, int K) {
        stack<string> st;
        string line;
        for(int i=S.length()-1;i>=0;i--){
            if(S[i]=='-') continue;
            
            if(line.length()<K){
                if(S[i]>='a' && S[i]<='z') line.insert(0, 1, S[i]-'a'+'A');
                else line.insert(0, 1, S[i]);
                
                if(line.length()==K){
                    st.push(line);
                    line="";
                }
            }
        }
        if(st.empty()) return line;
        
        if(line.length() == 0){
            line += st.top();
            st.pop();
        }
        while(!st.empty()){
            line+='-';
            line+=st.top();
            st.pop();
        }
        return line;
    }
};