题解 | #字符串合并处理#

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);

        while(in.hasNext()){
            solution(in);
        }
    }

    /**
     * 模拟法
     * @param in
     */
    private static void solution(Scanner in){

        // 1. 前后合并
        String str = in.nextLine().replaceAll(" ", "");

        // 2. 奇偶排序
        StringBuilder odd = new StringBuilder();
        StringBuilder even = new StringBuilder();
        for(int i=1; i<=str.length(); i++){
            char ch = str.charAt(i-1);
            if(i%2 == 0){
                even.append(ch);
            }else{
                odd.append(ch);
            }
        }

        char[] oddArray = odd.toString().toCharArray();
        char[] evenArray = even.toString().toCharArray();

        Arrays.sort(oddArray);
        Arrays.sort(evenArray);
        
        // 3. 字符转换
        int oddLen = oddArray.length;
        int evenLen = evenArray.length;
        StringBuilder result = new StringBuilder();
        for(int i=0; i<evenLen; i++){
            result.append(transfer(oddArray[i]));
            result.append(transfer(evenArray[i]));
        }

        // 特殊情况 奇偶不等
        if(oddLen > evenLen){
            result.append(transfer(oddArray[oddLen-1]));
        }

        System.out.println(result);
    }

    /**
     * 字符转换
     * @param ch
     * @return
     */
    private static Character transfer(char ch){
        if(String.valueOf(ch).matches("[0-9A-Fa-f]")){
            // 16进制 -> 10进制 -> 2进制
            String chBinary = String.format("%04d", Integer.parseInt(Integer.toBinaryString(Integer.parseInt(String.valueOf(ch), 16))));
            // 倒序
            String chReverse = new StringBuilder(chBinary).reverse().toString();

            // 2进制 -> 10进制 -> 16进制(大写)
            return Integer.toHexString(Integer.parseInt(chReverse, 2)).toUpperCase().charAt(0);
        }

        return ch;
    }
}