题解 | #字符串合并处理#

字符串合并处理

https://www.nowcoder.com/practice/d3d8e23870584782b3dd48f26cb39c8f

import java.util.*;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s1 = scanner.next();
        String s2 = scanner.next();
        scanner.close();
        //步骤1,合并
        String s3 = s1 + s2;
        //步骤2,分两个,然后排序
                StringBuilder stringBuilderS1 = new StringBuilder();//奇
        StringBuilder stringBuilderS2 = new StringBuilder();//偶
        for (int i = 0; i < s3.toCharArray().length; i++) {
            if (i % 2 == 0) {
                stringBuilderS1.append(s3.toCharArray()[i]);
            } else {
                stringBuilderS2.append(s3.toCharArray()[i]);
            }
        }
        char[] charArray1 = stringBuilderS1.toString().toCharArray();
        Arrays.sort(charArray1);
        char[] charArray2 = stringBuilderS2.toString().toCharArray();
        Arrays.sort(charArray2);
        //变为新的字符串
        char[] chars = new char[s3.length()];
        for (int i = 0; i < s3.length(); i++) {
            if (i % 2 == 0) {
                chars[i] = charArray1[i / 2];
            } else {
                chars[i] = charArray2[i / 2];
            }
        }
        //第三步,加密
        for (int i = 0; i < chars.length; i++) {
            char aChar = chars[i];
            //需要转化的部分
            if (Character.isDigit(aChar) || aChar >= 'a' && aChar <= 'f' || aChar >= 'A' &&
                    aChar <= 'F') {
                int tenAchar = Character.digit(aChar, 16); //转化为10进制
                int reverse = Integer.reverse(tenAchar) >>> (32 -
                              4); //reverse:反转,但只针对32位,需要取4位,所以→偏移32-4
                String string = Integer.toString(reverse, 16);//把这个数字转化为16进
                aChar = Character.toUpperCase(string.charAt(0));
            }
            System.out.print(aChar);
        }
    }
}

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