Huawei written test exercise ---- parses 9-in-order, and the 11-in-order string is a 10-in-order digital output.

Parses the 9-in-hexadecimal format, and the 11-in-hexadecimal string is a 10-in-hexadecimal numeric output.
Description:	Parse the 9-in-9 format. The 11-in-11 string is a 10-in-10 numeric output. The length of the input string's valid bits (0v0012345678) is no more than 8 bits. The previous 00 is not a valid bits. After resolution, it is output in decimal format. -1 is returned if an invalid string is parsed. 9 hexadecimal: The value range of the 9-digit number is 0, 1, 2, 3, 4, 5, 6, 7, and 8. 9: 0 V or 0 V Example of a 9-digit system: 0v11 0v564 0v123 0v0784 0v0 0 V 0 V 9-digit error instance: 0v923 0vt12 00v21 0123 9-to-10: 0v11-> 10 0v564-> 463   11 hexadecimal: The range of numbers in the 11 hexadecimal format: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A () The start of the 11 hexadecimal format is 0 w or 0 w. Correct 11-digit example: 0w11 0w564 0w123 0w0a8a 0 w 0 w 11 hexadecimal error instance: 0wb923 0 wvaa 00w21 0wax123 Convert the hexadecimal value to the hexadecimal value: 0w11-> 12 0w564-> 675
Running time limit:	Unlimited
Memory limit:	Unlimited
Input:	The input is a line of 9-or 11-digit strings. The format is shown in the preceding figure.
Output:	Number in decimal format
Sample input:	0w564
Sample output:	675
Answer:

I implemented this question in Java, because the 9-and 11-digit strings have certain rules, so I want to use regular expressions for matching. The 9-digit regular expression is ^ 0 [v | V] [0-8] * $, the Regular Expression in decimal format is ^ 0 [w | w] [0-9 | A] * $. If the input string matches the format, it is parsed. Otherwise, enter-1.

Code implementation:

import java.util.Scanner;  import java.util.regex.Pattern;    public class Main {            public int charge(String num){          int res=0;          String reg1 = "^0[v|V][0-8]*$";          String reg2 = "^0[w|W][0-9|a|A]*$";          char sub[];          if(Pattern.matches(reg1, num)){              sub = num.substring(2).toCharArray();              for(int i = 0;i< sub.length; i++){                  int count =Integer.parseInt(String.valueOf(sub[i]));;                  res += count * Math.pow(9, sub.length-i-1);              }              return res;          }else if(Pattern.matches(reg2, num)){              sub = num.substring(2).toCharArray();              for(int i = 0;i< sub.length; i++){                  if(sub[i]==‘a‘||sub[i]==‘A‘){                    res += 11 * Math.pow(11, sub.length-i-1);                  }else {                    int count = Integer.parseInt(String.valueOf(sub[i]));                    res += count * Math.pow(11, sub.length-i-1);                  }              }              return res;          }          return -1;      }      public static void main(String[] args) {              Scanner cin = new Scanner(System.in);              Main main = new Main();              String m;              while (cin.hasNext()){                  m = cin.next();                   System.out.println(main.charge(m));              }  

This code runs in IDE. However, in a bad test environment of Huawei, a test case fails. Because the test case is not made public, I have not found the reason, I hope users can help find bugs.

Huawei written test exercise ---- parses 9-in-order, and the 11-in-order string is a 10-in-order digital output.