有没有更好的方法可以用Java编写呢？

Question

public class Palindrome
{
   public static void main ( String [] args )
   {
       Scanner scan = new Scanner ( System.in ) ; // new object named "scan"
       System.out.println ( "\nPlease enter a number: "  ) ;
       int num = scan.nextInt () ; //store number into num

       if ( ( (num /1000)==(num%10) ) 
            && (    ( (num/10) %10 ) == ( (num/100)%10  )   ) ) 
           // num = 'abcd' : if (a=d) and (b=c) number is a palindrome.
            System.out.println ( "\nIs a Palindrome !"  ) ;
      else 
            System.out.println ( "\nNot a Palindrome !"  ) ;
   } //end of method main
} //end of class Palindrome

1. 有没有一种方法可以优化此代码？ 我只需要处理一个4位数字！
2. 如何添加数字是否为正数和4位数字的校验？

Answer 1

将任务分解为更小，更易于管理的部分，您可以解决。 例如：

要检查字符串的长度是否为4：

static boolean is4Digit(String p) {  
  return p.length()==4; 
} 

要检查数字是否在2个数字之间：

static boolean isBetween(int nr, int low, int hi) {  
  return nr>low && nr<hi; 
} 

要检查字符串是否是回文：

static boolean isPalindrome(String p) {  
  return p.equals(new StringBuilder(p).reverse().toString());  
}  

Answer 2

如何添加一个数字，以确保它是真正的数字：第一个是正数，第二个是4位数字？

我可以用两种方式解释4位非负整数。

0 <= n && n <= 9999

要么

1000 <= n && n <= 9999

为了进行优化，我不会从字符串中解析它：

BufferedReader bufferedInput = new BufferedReader(System.in);
String line = bufferedInput.readLine();
if (line == null) { /* handle end of input case */ }
String number = line.trim();
// Check if its a palindrome
if (number.length() == 4
    && number.charAt(0) == number.charAt(3)
    && number.charAt(1) == number.charAt(2)) {
  // Finally check that the characters are all digits
  boolean isLegal = true;
  for (int i = (number.length() + 1) / 2; --i >= 0;) {
    char shouldBeADigit = number.charAt(i);
    if (!('0' <= shouldBeADigit && shouldBeADigit <= '9')) {
      isLegal = false;
      break;
    }
  }
  if (isLegal) {
    System.out.println(number + " is a palindrome");
  }
}

Answer 3

更具可读性的方式是：

   Scanner scan = new Scanner ( System.in ) ; // new object named "scan"
   System.out.println ( "\nPlease enter a number: "  ) ;
   int num = scan.nextInt () ; //store number into num

   if(num<0 || num>9999)
       System.out.println("Only positive numbers with 4 digits are accepted.");
   else {
       int help=num;
       int rev=0;

       while (help > 0) {
           rev = rev * 10 + help %10;
           help = help / 10;
       }
       System.out.println("Is a palindrome: "+ (rev==num));
   }

Answer 4

检查输入在哪里是数字

public static boolean isNumeric(String str)
{
  NumberFormat formatter = NumberFormat.getInstance();
  ParsePosition pos = new ParsePosition(0);
  formatter.parse(str, pos);
  return str.length() == pos.getIndex();
}

Answer 5

isPalindrome的纯数学实现（无字符串转换）比Chris的效率更高（尽管当然甚至不那么直观/可读）：

public static boolean isPalindrome(int number) {
  if(number < 0){
    throw new IllegalArgumentException("number must be positive!");
  }
  int powerOf10 = (int)Math.pow(10, (int)Math.floor(Math.log10(number)));
  while(powerOf10 >= 10) {
    number -= powerOf10 * (number % 10);
    if(number < 0 || number >= powerOf10) {
      return false;
    }
    number /= 10;
    powerOf10 /= 100;
  }
  return true;
}

由于这是一个家庭作业问题，因此我将向读者说明为什么这样做是一种练习。

Answer 6

您也可以使用带有捕获组和反向引用的正则表达式进行此操作。 考虑此正则表达式以捕获两个数字，然后以相反的顺序捕获另外两个与前两个数字相等的数字：

 private static String PALINDROME = "(\\d)(\\d)(?:\\2)(?:\\1)";

 public boolean isPalindrome(String testStr) throws NoMatchFoundException {
     Pattern pattern = Pattern.compile(PALINDROME);
     Matcher matcher = pattern.matcher(testStr);
     return matcher.matches(); 
 }