sms_problem.txt

/*
 SMS Problem
 0 - NULL, 1 - NULL, 2 - ABC, 3 - DEF, 4 - GHI, 5 - JKL, 6 - MON, 7 - PQRS, 8 - TUV, 9 - WXYZ, * - <Space>, # - <Break>
 We must convert the numbers to text.
 Eg: 22 - B, 23 - AD, 223 - BD, 22#2 - BA (# breaks the cycle), 3#33 - DE, 2222-22222#2 - 2A, 22222 - A (cycle must wrap around), 222222 - B
*/

#include <iostream>
using namespace std;

int main(){
    cout<<"input a string of number"<<endl;
    string str;
    cin>>str;
    /*Verify the input*/
    for (int i=0; i<str.length(); ++i){
        if (!((str[i]<='9' && str[i]>='0')||str[i]=='#'||str[i]=='*')){
            cout<<"wrong input!"<<endl;
            return -1;
        }
    }
    
    string keypad[] = {"", "", "ABC2", "DEF3", "GHI4", "JKL5", "MON6", "PQRS7", "TUV8", "WXYZ9"}; // note: start from '2', add number at the end of array.
    int i=0, j=0, n = str.length();//n is the length of the original string.
    while(i<n){
        if (str[i]=='#'){
            i++;
            continue;
        }
        else if (str[i]=='*'){
            str[j++]=' ';
            i++;
            continue;
        }
        else{
            int count = 0;
            while(str[i]==str[i+1]){
                if (i==n-1){
                    count++;
                    break;
                }
                count++;
                i++;
                
            }
            str[j]=keypad[str[i]-'0'][(count)%(keypad[str[i]-'0'].length())];
            j++;
            i++;
        }
    }
    cout<<"The sms is ";
    for (int i=0; i<j; ++i)
        cout<<str[i];
    cout<<endl;
    return 0;
}

REMARK:
1. The idea is pretty straightforward. In the outer while loop, you just need to see which case is in: a#, a* or a digit.