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Java-based LeetCode algorithm problem solutions, regularly updated
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9\. Palindrome Number
Easy
Given an integer `x`, return `true` if `x` is palindrome integer.
An integer is a **palindrome** when it reads the same backward as forward. For example, `121` is palindrome while `123` is not.
**Example 1:**
**Input:** x = 121
**Output:** true
**Example 2:**
**Input:** x = -121
**Output:** false
**Explanation:** From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
**Example 3:**
**Input:** x = 10
**Output:** false
**Explanation:** Reads 01 from right to left. Therefore it is not a palindrome.
**Example 4:**
**Input:** x = -101
**Output:** false
**Constraints:**
* -231 <= x <= 231 - 1
**Follow up:** Could you solve it without converting the integer to a string?
To solve the Palindrome Number problem in Java using a `Solution` class, we'll follow these steps:
1. Define a `Solution` class with a method named `isPalindrome`.
2. Handle special cases where `x` is negative or divisible by 10 but not equal to zero, as they cannot be palindromes.
3. Initialize variables to keep track of the reversed number (`reversed`) and the original number (`original`).
4. Iterate through the digits of the original number `x`:
- Extract the least significant digit using the modulo operator and append it to the reversed number.
- Update `original` by removing the least significant digit using integer division.
5. Check if the reversed number is equal to the original number. If so, return `true`; otherwise, return `false`.
Here's the implementation:
```java
public class Solution {
public boolean isPalindrome(int x) {
// Special cases:
// As discussed, negative numbers are not palindromes.
// Also, if the last digit of the number is 0, it can't be a palindrome unless the number is 0 itself.
if (x < 0 || (x % 10 == 0 && x != 0)) {
return false;
}
int reversed = 0;
int original = x;
while (original > reversed) {
int digit = original % 10;
reversed = reversed * 10 + digit;
original /= 10;
}
// When the length is an odd number, we can get rid of the middle digit by reversed / 10
// For example when the input is 12321, at the end of the while loop we get x = 12, reversed = 123,
// since the middle digit doesn't matter in palidrome(it will always equal to itself), we can simply get rid of it.
return original == reversed || original == reversed / 10;
}
public static void main(String[] args) {
Solution solution = new Solution();
// Test cases
int x1 = 121;
System.out.println("Example 1 Output: " + solution.isPalindrome(x1));
int x2 = -121;
System.out.println("Example 2 Output: " + solution.isPalindrome(x2));
int x3 = 10;
System.out.println("Example 3 Output: " + solution.isPalindrome(x3));
int x4 = -101;
System.out.println("Example 4 Output: " + solution.isPalindrome(x4));
}
}
```
This implementation provides a solution to the Palindrome Number problem in Java.
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