9. Palindrome Number
Easy
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Example 1:
Input: 121 Output: true
Example 2:
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.
Example 3:
Input: 10 Output: false Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Follow up:
Coud you solve it without converting the integer to a string?
package leetcode.easy;
public class PalindromeNumber {
@org.junit.Test
public void test() {
int number1 = 121;
int number2 = -121;
int number3 = 10;
PalindromeNumber palindromeNumber = new PalindromeNumber();
System.out.println(palindromeNumber.isPalindrome(number1));
System.out.println(palindromeNumber.isPalindrome(number2));
System.out.println(palindromeNumber.isPalindrome(number3));
}
public boolean 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;
}
}