DevExplained

7. Reverse Integer

Math

Explanation

To reverse an integer, we can use the following steps:

  1. Initialize a variable reversed to store the reversed number.
  2. Handle negative numbers by checking if the input is negative and then multiplying the reversed number by -1 at the end.
  3. Iterate through each digit of the input number by continuously dividing it by 10 and extracting the last digit.
  4. Multiply the current reversed number by 10 and add the extracted digit to the ones place.
  5. Check for overflow by comparing the new reversed number with the integer limits.
  6. Return the final reversed number.

The time complexity of this approach is O(log(x)) where x is the input number, as we iterate through the digits of the number. The space complexity is O(1) as we only use a constant amount of extra space.

View Leetcode Problem
class Solution {
    public int reverse(int x) {
        int reversed = 0;
        while (x != 0) {
            int digit = x % 10;
            x /= 10;
            if (reversed > Integer.MAX_VALUE / 10 || (reversed == Integer.MAX_VALUE / 10 && digit > 7)) return 0;
            if (reversed < Integer.MIN_VALUE / 10 || (reversed == Integer.MIN_VALUE / 10 && digit < -8)) return 0;
            reversed = reversed * 10 + digit;
        }
        return reversed;
    }
}

Code Editor (Testing phase)

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