判断回文数

力扣

package Algorithms;

/**
 * @author : zhang
 * @version : 1.0
 * @date : Create in 2021/7/28
 * @description :
 */
public class HuiWen {
    public static void main(String[] args) {
        int x = 100001;
        boolean palindrome = isPalindrome_2(x);
        System.out.println(palindrome);
    }

    //解法1，转化为字符串，反转后和原字符串进行比较
    public static boolean isPalindrome_1(int x) {
        String reversedStr = (new StringBuilder(x + "").reverse().toString());
        return (x + "").equals(reversedStr);
    }

    //解法2，数字反转后和原数字进行比较
    public static boolean isPalindrome_2(int x) {
        if (x<0) return false;
        int reverseNumber = 0;
        int num = x;
        while (num!=0){
            reverseNumber = reverseNumber*10 +num %10;
            num/=10;
        }
        return reverseNumber == x;
    }

    //解法3，取出后半数，和前半数进行比较
    public static boolean isPalindrome_3(int x) {
        if (x < 0 || (x % 10 == 0 && x != 0)) return false;
        int revertedNumber = 0;
        while (x > revertedNumber) {
            revertedNumber = revertedNumber * 10 + x % 10;
            x /= 10;
        }
        return x == revertedNumber || x == revertedNumber / 10;
    }

}