PalindromeNumber.java

class Solution {
    /*Determine whether an integer is a palindrome. 
    An integer is a palindrome when it reads the same backward as forward.
    
    Input: -121
    Output: false
    Explanation: From left to right, it reads -121. From right to left, it becomes 121-. 
    Therefore it is not a palindrome
    */
    public static void main(String[] args) {
        int x = -121;
        if(isPallindrome(x))
            System.out.println("TRUE");
    }

    public static boolean isPallindrome(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!=0 && x%10==0))
            return false;
        
        int revertedNumber = 0;

        while(x > revertedNumber){
            revertedNumber = revertedNumber*10 + x%10;
            x = 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);
        
    }
    /*Time complexity : O(\log_{10}(n))O(log10(n)).
     We divided the input by 10 for every iteration, 
     so the time complexity is O(\log_{10}(n))O(log10(n))
    Space complexity : O(1)O(1).*/
}