力扣-回文数

• 我写的

    public static boolean isPalindrome(int x) {
        if (x < 0 || (x % 10 == 0 && x != 0)) {
            return false;
        }
        String str = Long.toString(x);
        String str1 = str.substring(0, str.length() / 2);
        String str2 = new StringBuffer(str).reverse().toString().substring(0, str.length() / 2);
        return str1.equals(str2);
    }

• 官方

  public boolean isPalindrome(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;
    }