Determine whether an integer is a palindrome. Do this without extra space.

It turns out that there is no need to prevent overflow when reversing the integer. Palindrome of X will result in a negative number because of overflow, and that makes it never equals to the input number x

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public class Solution {
    public boolean isPalindrome(int x) {
        //if (x < 0 || x == Integer.MAX_VALUE){return false;}
        
        int old = x, cur = 0;
        
        while (x > 0){
            cur = cur * 10 + x % 10;
            x /= 10;
        }
        
        return old == cur;
    }
}

A solution that prevents overflow.

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public class Solution {
	public boolean isPalindrome(int x) {
        if (x != 0 && x % 10 == 0) return false;
        int sum = 0;
        
        // half of the digits
        while (x > sum) {
            sum = sum * 10 + x % 10;
            x /= 10;
        }
        // even || odd 
        return (x == sum) || (x == sum / 10);
    }
}