第七个怎么写？求解答

def is_palindrome(num_s):
    flag = True
    length = len(num_s)
    if  length== 1:
        return flag

    for i in range(0, int(length/2)):
        if num_s[i] != num_s[length - i - 1]:
            flag = False
            break
    return flag

# 键盘输入k
k = int(input("请输入k值："))
print(is_palindrome(k))

import java.util.Scanner;
public class Number {
   public static void main(String args[]) {
      int number=0,d5,d4,d3,d2,d1;
      Scanner reader = new Scanner(System.in);
      System.out.println("输入一个1至9999之间的数");
      number = reader.nextInt();
      if(number>=1&&number<=9999) //判断number在1至9999之间的条件
      {        
               d4=number%10000/1000;   //计算number的千位d4
                   d3=number%1000/100 ;  //计算number的百位d3
                d2=number%100/10;
                d1=number%10;
                if(d4!=0)  //判断number是4位数的条件
                {  System.out.println(number+"是4位数");
                   if(d3==d2&&d4==d1) //判断number是回文数的条件码
                   {  System.out.println(number+"是回文数");
                   }
                  else
                   {  System.out.println(number+"不是回文数");
                   }
                }
                else if(d3!=0)  //判断number是3位数的条件
                {  System.out.println(number+"是3位数");
                   if(d1==d3) //判断number是回文数的条件
                   {  System.out.println(number+"是回文数");
                   }
                  else
                   {  System.out.println(number+"不是回文数");
                   }
                }
                else if(d2!=0)
                {  System.out.println(number+"是2位数");
                   if(d1==d2)
                   {  System.out.println(number+"是回文数");
                   }
                  else
                   {  System.out.println(number+"不是回文数");
                   }
                }
                else if(d1!=0)
                {  System.out.println(number+"是1位数");
                   System.out.println(number+"是回文数");
                }
            }
      else 
      {  System.out.printf("\n%d不在1至99999之间",number);
      }
   }
}

用闸瓦写的，看看行不行呀