Interview Question Reverse Integer

By Fatih KABAKCI

The problem is to reverse an integer that is entered by the user. The main points of the solution are to check last and first characters. First, if the last character is already 0, then the result should be 0 too. On the other hand, first character might be entered by adding either + or - operator. In this case, first character is ignored when the integer is being converted to the string. The basic solution has been shared the following.

Problem #7 - Reverse Integer

Description: Reverse digits of an integer.

Difficulty: Easy

Solution in Java

package com.fatihkabakci.Easy.ReverseInt;

/**
 * 
 * @author fkabakci
 * Problem Description: Reverse digits of an integer.
 * 
 * Examples:
 * 
 * Example1: x = 123, return 321
 * 
 * Example2: x = -123, return -321
 * 
 * Solution:
 * 
 * 1. Convert it to String.
 * 2. Check the last character if it is 0, if so return 0.
 * 3. Check the first character if it has + or - operator, if so, then cut the first one in string, 
 *    and specify it positive or negative.
 * 4. Reverse all characters
 * 5. Convert it to Integer, then return the result. If overflow occurs return 0.
 */
public class ReverseInteger {
	public int reverse(int x) {
		String sx = String.valueOf(x);
		// last character if 0 ?
		char cn = sx.charAt(sx.length() - 1);
		if (cn == "0")
			return 0;

		short pre = 1;
		// first character + or - ?
		char c0 = sx.charAt(0);
		if (c0 == "+") {
			pre = 1;
			sx = sx.substring(1);
		} else if (c0 == "-") {
			pre = -1;
			sx = sx.substring(1);
		}

		char[] cx = sx.toCharArray();
		char[] rx = new char[cx.length];
		for (int i = cx.length - 1; i >= 0; i--) {
			// reversing
			rx[cx.length - 1 - i] = cx[i];
		}
		try {
			int result = Integer.valueOf(new String(rx));

			return pre * result;
		} catch (NumberFormatException nfe) {
			return 0;
		}
	}

	public static void main(String[] args) {
		ReverseInteger r = new ReverseInteger();
		System.out.println(r.reverse(123));
		System.out.println(r.reverse(+123));
		System.out.println(r.reverse(-123));
		System.out.println(r.reverse(1534236469));
	}
}

The algorithm complexity is O(n) because of n character.