Solution by ccwei

  • 3

    Approach #1 Revert half of the number [Accepted]


    The first few idea comes to mind is to convert the number into string, and check if the string is a palindrome, but this would require extra none-constant space for creating the string which is not allowed by the problem description.

    Second idea would be reverting the number itself, and then compare the number with original number, if they are the same, then the number is a palindrome. However, if the reversed number is larger than Int.Max, we will hit integer overflow problem.

    Follows the thoughts on the second idea, to avoid the overflow issue of the reverted number, what if we only revert half of the int number? After all, the reverse of the last half of the palindrome should be the same as the first half of the number, if the number is a palindrome.

    For example, if the input is 1221, if we can revert the last part of the number "1221" from "21" to "12", and compare with the first half of the number "12", since 12 is the same as 12, we know the number is a palindrome.

    Let's see how we could translate this idea into an algorithm.


    First of all we should take care of some edge cases. All negative numbers are not palindrome, for example: -123 is not a palindrome since the '-' does not equal to '3'. So we can return false for all negative numbers.

    Now let's think about how to revert the last half of the number. For number 1221, if we do 1221 % 10, we get the last digit 1, to get the second to the last digit, we need to remove the last digit from 1221, we could do so by divided it by 10, 1221 / 10 = 122. Then we can get the last digit again by %10, 122 % 10 = 2, and if we time the first last digit by 10 and plus the second last digit, 1 * 10 + 2 = 12, it gives us the reverted number we want. Continue this process would give us reverted number with more digits. Now the question is, how do we know that we reaches the half of the number?

    Since we divided the number by 10, and multiple the reversed number by 10, when the original number is less than the reversed number, it means we've processed half of the number digits.


    public class Solution {
        public bool IsPalindrome(int x) {
            // Special cases:
            // As discussed above, when x < 0, x is not a palindrome.
            // Also if the last digit of the number is 0, in order to be a palindrome, 
            // the first digit of the number also needs to be 0.
            // Only 0 satisfy this property.
            if(x < 0 || (x % 10 == 0 && x != 0)) {
                return false;
            int revertedNumber = 0;
            while(x > revertedNumber) {
                revertedNumber = revertedNumber * 10 + x % 10;
                x /= 10;
            // When the length is an odd number, we can get rid of the middle digit by revertedNumber/10
            // For example when the input is 12321, at the end of the while loop we get x = 12, revertedNumber = 123, 
            // since the middle digit doesn't matter in palidrome(it will always equal to itself), we can simply get rid of it.
            return x == revertedNumber || x == revertedNumber/10;

    Complexity Analysis

    • Time complexity : $$O({log n})$$.
      We divided the input by 10 for every iteration, so the time complexity is $$O({log_10 n})$$

    • Space complexity : $$O(1)$$.

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