[PAT] 1023 Have Fun with Numbers （20 分）Java

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

Sample Input:

1234567899

Sample Output:

Yes
2469135798

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;

/**
 * @Auther: Xingzheng Wang
 * @Date: 2019/2/21 18:07
 * @Description: pattest
 * @Version: 1.0
 */
public class PAT1023 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        BigInteger number = sc.nextBigInteger();
        BigInteger result = number.multiply(new BigInteger("2"));
        ArrayList<Character> list1 = new ArrayList();
        ArrayList<Character> list2 = new ArrayList();
        sortOfNumber(list1,number);
        sortOfNumber(list2,result);

        if(list1.equals(list2))
            System.out.println("Yes");
        else
            System.out.println("No");
        System.out.println(result);
    }
    private static void sortOfNumber(ArrayList<Character> list,BigInteger number){
        String s = number.toString();
        char[] ch = s.toCharArray();
        for (char c : ch){
            list.add(c);
        }
        Collections.sort(list);
    }
}