Converting int value to String without using toString and parseInt method

I am studying about the converting int to string and string to int without using toString and parseInt method. I already made methods. However, I don't believe that my way of doing was the best. I would like to improve these function.

For example, the first converts from String to int:

public static int StringToint(String number)
        int eachnumber = 0;
    int intConvert = 48;
    int reVal = 0;
    int index;
    int maxlen =number.length() - 1;
        for(index = 0 ; index <= maxlen ; index++)
    {
        eachnumber = number.charAt(index);

        eachnumber = eachnumber  - intConvert;

        reVal = reVal + (eachnumber * (int) Math.pow(10, maxlen - index));
    }
    return reVal;
}

Second, for converting int to String:

public static String IntToString(int number)
{
    int StringConvet = 48;

    int eachDigit = number;
    int afterDivide = number;
    String reVal = "";

    while(afterDivide >0)
    {
        eachDigit = afterDivide % 10;
        afterDivide = afterDivide / 10;
        if(eachDigit == 0)
        {
            reVal += "0";
        }
        else if(eachDigit == 1)
        {
            reVal += "1";
        }
        else if(eachDigit == 2)
        {
            reVal += "2";
        }
        else if(eachDigit == 3)
        {
            reVal += "3";
        }
        else if(eachDigit == 4)
        {
            reVal += "4";
        }
        else if(eachDigit == 5)
        {
            reVal += "5";
        }
        else if(eachDigit == 6)
        {
            reVal += "6";
        }
        else if(eachDigit == 7)
        {
            reVal += "7";
        }
        else if(eachDigit == 8)
        {
            reVal += "8";
        }
        else if(eachDigit == 9)
        {
            reVal += "9";
        }
    }
    String reVal2 = "";
    for(int index =  reVal.length() -1 ; index >= 0 ; index--)
    {
        reVal2 += reVal.charAt(index);
    }
    return reVal2;
}

I am sure that t

To improve your intToString() method you should consider using a StringBuilder, and specifically the method StringBuilder.append(int).

Iterate digits in your int, and for each digit you can append(eachDigit) to the StringBuilder element. This will also reduce the complexity of intToString() to \$O(n)\$ since you do not need to create a new String instance each iteration. To get a String object from the StringBuilder, use StringBuilder.toString(). Or, if you are not allowed, you can use StringBuilder.subString(0).

You should also use a StringBuilder.append() (using the same idea) to reverse the resulting string (your second loop in your code).

Since it is not homework (as per comments), I have no problems providing a code snap. It should look something like this:

public static String intToString(int n) { 
    if (n == 0) return "0";
    StringBuilder sb = new StringBuilder();
    while (n > 0) { 
        int curr = n % 10;
        n = n/10;
        sb.append(curr);
    }
    String s = sb.substring(0);
    sb = new StringBuilder();
    for (int i = s.length() -1; i >= 0; i--) { 
        sb.append(s.charAt(i));
    }
    return sb.substring(0);
}

Notes:

• You can also use StringBuilder.reverse() instead of the second loop.

• In here, \$O(n)\$ means linear in the the number of digits in the input number (n is the number of digits in the input number - not the number itself!) If you are looking for the complexity in terms of the initial number (it is \$O(\log(n))\$) since you divide your element by 10 each iterations, you have a total of \$\log_{10}(\text{number})\$ iterations for each loop, which results in \$O(\log(\text{number}))\$.

StackExchange Code Review Q#10351, answer score: 12