题解 | #字符串合并处理#-Java

import java.util.*;
public class Main{
    private static char[] array={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        while(input.hasNext()){
            String str1 = input.next();
            String str2 = input.next();
            String mergeStr = str1 + str2;
            //长度为奇数时，偶数序列多一个
            int len1 = mergeStr.length() % 2 == 0 ? mergeStr.length() / 2 : mergeStr.length() / 2 + 1;
            int len2 = mergeStr.length() / 2;
            char[] array1 = new char[len1];
            char[] array2 = new char[len2];
            for(int i = 0;i < mergeStr.length();i++){
                if(i % 2 == 0){
                    array1[i / 2] = mergeStr.charAt(i);
                }else{
                    array2[i / 2] = mergeStr.charAt(i);
                }
            }
            Arrays.sort(array1);
            Arrays.sort(array2);
            StringBuilder builder = new StringBuilder();
            for(int i = 0;i < mergeStr.length(); i++){
                if(i % 2 == 0){
                 builder.append(bitRotate(array1[i / 2]));   
                }else{
                 builder.append(bitRotate(array2[i / 2])); 
                }
            }
            System.out.println(builder.toString());
        }
    }

    //这里反转简单思路是例 0001 是从右到左2 bit下标对应n次之和，翻转就从左到右2的n次之和就可以
    private static char bitRotate(char a){
        int number = 0;
        if(a >= 'a' && a<= 'f'){
            number = a - 'a' + 10;
        }else if(a >= 'A' && a <= 'F'){
            number = a - 'A' + 10;
        }else if(a >= '0' && a <= '9'){
            number = a - '0';
        }else{
            return a;
        }
        int tag = 8;
        int result = 0;
        for(int i = 0;i<4;i++){
            if((number & tag) == tag){
                result += Math.pow(2,i);
            }
            number = number << 1;
        }
        return array[result];
    }
}