CareerCup

Java:

public class IntegerToString {

	public static void main(String[] args) {
		System.out.println(itoa(4973));
		System.out.println(itoa(-59385));
		System.out.println(itoa(0));
	}

	private static String itoa(int num) {
		if(num == 0){
			return "0";
		}
		boolean negative = num < 0;
		String result = "";
		while(num != 0) {
			result = Math.abs(num%10) + result;
			num /= 10;
		}
		return negative ? "-" + result : result;
	}

}

Outputs:

4973
-59385
0

It's actually quite simple. Use mod and div.

Use recursion

char result [50]={'0'};
int i=0;

itoa(int a){
if(!a) return;
else{
itoa(a/10);
result[i++]=a%10+'0';
}
}

Do checks on negative numbers before and increment i by 1 storing '-' at 0th position of result

Use recursion

char result [50];
int i=0;

void itoa(int a){
if(!a) return;
else 
itoa(a/10);
result [i++]= a%10+'0';
}

Check for negative and zero before calling recursion and make changes in the result array accordingly.

How about this? Nice and slim!

System.out.println("an itoa");

public class Solution1 {
	public static void main (String args[]) {
		Solution1 s = new Solution1();
		System.out.println(s.solve(-333));
	}
	public String solve(int number) {
		if (number == 0)
			return "0";
		String results = "";
		boolean negative = false;
		if (number < 0){
			results = "-";
			negative = true;
		}
		if (negative)
			number = number * (-1);
		int remain = number % 10;
		number = number / 10;
		if (number != 0)
			return results+""+solve(number)+remain;
		return results+""+remain;
	}
}