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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 the StringToint method is \$O(n)\$ and intToString method will be \$O(n^2)\$. However, I am really sure that there are ways to improve these two methods. Is there any way to improve these two functions?

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migrated from Mar 26 '12 at 21:50

This question came from our site for professional and enthusiast programmers.

Your making these on purpose right.. Instead of just using Integer.parseInt() and Integer.toString().. – barsju Mar 26 '12 at 21:35
yep, I am studying about interview questions and I just want to know the better way to build these methods. – Dc Redwing Mar 26 '12 at 21:41
What happens if you need a string representation of a negative number? – Clockwork-Muse Apr 4 '12 at 21:56
For the love of baby Jesus, fix your indentation! – Emily L. Jul 14 '15 at 15:18
up vote 9 down vote accepted

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 to provide a code snap. It should look something like that:

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;
    String s = sb.substring(0);
    sb = new StringBuilder();
    for (int i = s.length() -1; i >= 0; i--) { 
    return sb.substring(0);

(*) Note you can also use StringBuilder.reverse() instead of the second loop.

(*)Also note: 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 - and not the number itself!] If you are looking for the complexity in terms of the initial number - it is O(logn) - since you divide your element by 10 each iterations, so you have total of log_10(number) iterations for each loop, which results in total of O(log(number)).

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sorry , it is not homework. It actually came from the interview questions. I was preparing the interview for coding questions. Anyway, thanks a lot :) – Dc Redwing Mar 26 '12 at 21:42
This is still O(n^2) though, you can easily insert at beginning, otherwise there no need to use a StringBuilder (even because javac will optimize it to a StringBuilder in most cases) – Jack Mar 26 '12 at 21:54
@Jack: No, it is O(n) - each loop is O(n) - the algorithm is total of O(2 * n) = O(n). inserting element in the beginning will probably make it O(n^2) - since inserting an element to the beginning of dynamic array is O(n) each for each insert. – amit Mar 26 '12 at 21:58
Yes, yours it still O(2n), misread the code sorry. I was thinking about the internal implementation of StringBuilder, I thought it worked by doing some black magic with linked lists, not a plain internal array. But if it's a dynamic array of course insert will require O(n). I guess that with Math.ceil(Math.log10(number)) we could be able to avoid reversing the string then. – Jack Mar 26 '12 at 23:27

For your example , I should point out some problems : first , when you add two add string frequently, you should use StringBuilder instead ; second , you should consider Integer.MIN_VALUE into account!Here is my code:

public static String parseInt(int integer)
    boolean ifNegative = integer<0;
    boolean ifMin = integer == Integer.MIN_VALUE;
    StringBuilder builder = new StringBuilder();        
    integer = ifNegative?(ifMin?Integer.MAX_VALUE:-integer):integer;    
    List<Integer> list = new LinkedList<Integer>(); 
    int remaining = integer;
    int currentDigit = 0 ;

        currentDigit = remaining%10;
        remaining /= 10;
        if(remaining==0) break;

    currentDigit = list.remove(0);
    for(int c : list)
    builder.reverse().insert(0, ifNegative?'-':'+');
    return builder.toString();
share|improve this answer

Your approach seems indeed quite overkill.

Conversion from string to int can be done in a way similar to yours, but by multiplying the result by 10 at every step, without the need to calculate the i-th power for every digit:

int stringToInt(String s)
  int r = 0;
  for (int i = 0; i < s.length(); ++i)
    if (i > 0)
      r *= 10;

    r += s.charAt(i)-'0';
  return r;

While conversion from int to string can take into account the difference between the char '0' and the current character of the string without the need of a if/else chain. In addition you can use a string buffer to insert characters at the beginning to avoid the necessity of reversing the string.

String intToString(int i)
  StringBuilder b = new StringBuilder();

  while (i != 0)
    b.insert(0, (char)('0'+i%10));
    i /= 10;

  return b.toString();

Mind that the cast to char is needed to avoid using the method insert(int offset, int value) that would actually convert the int to a string.

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note that b.insert() is O(n). have a look on the performance of dynamic array [which I think this is how StringBuilder is implemented]. inserting an element to the beginning is O(n) each, so total of O(n^2). Also, there is a special case of i == 0, which should be handled. – amit Mar 26 '12 at 22:00

protected by Jamal Jul 14 '15 at 17:53

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