Reverse Integer

Table of contents
  1. Problem
  2. Explanation
  3. Solution

Problem

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-2^31, 2^31 - 1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Input: x = 123
Output: 321

Input: x = -123
Output: -321

Input: x = 120
Output: 21

Explanation

Everything else is pretty straightforward, but the tricky part is checking for overflow.

Because of the assumption, we cannot use a long to store the reversed integer, and then check if it is within the range of a 32-bit integer.

So I decided to store the reversed integer as a string, and then if we fail to parse it as an integer, we know it is out of range.

Solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

public int reverse(int x) {
  int sign = (x < 0) ? -1 : 1;
  x = sign * x;

  StringBuilder sb = new StringBuilder();

  while (x > 0) {
    sb.append(x % 10);
    x /= 10;
  }

  try {
    return sign * Integer.parseInt(sb.toString());
  } catch (NumberFormatException e) {
    return 0;
  }
}

The complexity is $O(\log x)$.