LeetCode 9: Palindrome Number Solution

Problem Explanation

Explanation:

To determine if a given integer is a palindrome without converting it to a string, we can reverse half of the number and compare it with the other half.

For negative numbers or numbers ending in 0 (except 0 itself), they cannot be palindromes.
Initialize a variable reversed to store the reversed half of the number.
While x is greater than reversed, reverse the last digit of x and add it to reversed.
If x has an odd number of digits, we can simply check if x == reversed/10.
If x has an even number of digits, we can check if x == reversed.

This approach avoids the need to convert the integer to a string.

Time complexity: O(log(x)) where x is the input number. Space complexity: O(1)

Solution Code

class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0 || (x % 10 == 0 && x != 0)) {
            return false;
        }
        
        int reversed = 0;
        while (x > reversed) {
            int lastDigit = x % 10;
            reversed = reversed * 10 + lastDigit;
            x /= 10;
        }
        
        return x == reversed || x == reversed / 10;
    }
}

Frequently Asked Questions

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The time complexity for LeetCode 9 (Palindrome Number) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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