Reverse Integer

Reverse digits of an integer. Returns 0 when the reversed integer overflows (signed 32-bit integer).

Example

Given x = 123, return 321

Given x = -123, return -321

Method 1: Use long type

Analysis

We need to deal with overflow very carefully.

One way is to use long type to explicitely check overflow by comparing with Integer.MAX_VALUE and Integer.MIN_VALUE.

Complexity

Time: O(logn)

Space: O(1)

Code

Java

public class Solution {
  private static final int BASE = 10;

  public int reverseInteger(int n) {
    long result = 0;
    while (n != 0) {
      result = result * BASE + n % BASE;
      n /= BASE;
    }
    return (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) ?
        0 : (int)result;
  }
}

Method 2: Use Integer

Analysis

Another way to do that is to check if the result before adding exceeds Integer.MAX_VALUE / BASE, where BASE = 10.

1. If exceeds, it overflow for sure after you add another value.
2. If not exceeds, it cannot overflow. Since the input argument's type is integer, so the new digit must be 1.

Code

Java

public class Solution {
  private static final int BASE = 10;
  private static final int THRESHOLD = Integer.MAX_VALUE / BASE;

  public int reverseInteger(int n) {
    int result = 0;
    while (n != 0) {
      if (Math.abs(result) > THRESHOLD) {
        result = 0;
        break;
      }
      result = result * 10 + n % BASE;
      n /= BASE;
    }
    return result;
  }
}