如何在单个方法中分别计算字符串中两个字符的出现？

Question

我的程序正在读取的字符串示例

[BABAAAABAAAAAAABAAAABBAAAAAABAAAABABAABAAABABABAABAAAAAABAAAAAABAAAAAA] >>

我目前有两个int数组int[] countA = new int[4]和int[] countB = new int[4]，其中每个数组中的四个元素对应于特定尺寸（Alpha，Bravo，Charlie，Delta）。我试图将上面的String分成7个组，其中第一个字符对应于Alpha，接下来的两个字符对应于Bravo，接下来的两个对应于Charlie，最后两个对应于Delta。如果字符是A或B，则会为相应的尺寸更新数字

（（在第9、10、16等字符处检测到B时，EX：countB[2]++] >>

目前，我的程序有两个单独的方法来计算两个字符的出现次数。我觉得这是一种多余的计数方式，因为两种方法中的代码几乎相同。我试图通过将两种计数方法压缩为一个方法来进一步优化程序。

当前程序代码：

import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintStream;
import java.util.Scanner;
import java.util.Arrays;
//CSC 142 , Check personality, personalityTest.java
// June 12, 2015, Chloe Wake
// Reads through a file of results from the Keirsey Personality test and outputs the answers in a output file.

public class CountDimensions {
   public static  final int DIMENSIONS = 4;//the number of personality dimensions

   public static void main(String[] args) throws FileNotFoundException{
      //Scanner console = new Scanner(System.in);
      userIntro(); //give an intro to the user
      processFile(); //read in the file and store it so we can access it
   }
   //method that reads in a file and stores each line in an array
   public static void processFile() throws FileNotFoundException  {
      Scanner console = new Scanner(System.in);//create scanner for user input
      File fileName =  inputFile(console); //store the file the user input typed in
      Scanner read1 = new Scanner(fileName);//create scanner for the .txt file specified by user
      Scanner inputFile = new Scanner(fileName); //make a new scanner to read the file now that we know it exists
      PrintStream output = new PrintStream(outputFile(console));//create printStream object for output
      // while loop to get the desired output
      while (inputFile.hasNextLine()) { // while the file has text in the line read through it
         String name = inputFile.nextLine(); // first line is the name of the person who took the test, go to the next line
         String  answers = inputFile.nextLine().toUpperCase(); // line after the name is the
         int[] findA = countA(answers);
         int[] findB = countB(answers);
         int[] percentage = percentage(findB);
         String[] type = type(percentage);
         outputFile(name, percentage, type, findA, findB);
      }

   }
   public static File inputFile(Scanner console)    {//ask for file
      System.out.print("Type the filename of the file you want to input: ");
      File answer = new File(console.nextLine());
      while (!answer.exists()) { //if it doesn't exist, throw an error
         System.out.println("That file does not exist! Try again.");
         answer = new File(console.nextLine()); //store the file the user input typed in
      }
      return answer;
   }
   //method that writes output back into a text file
   public static File outputFile(Scanner console)    {
      System.out.print("output file name? ");// PrintStream to write to a user specified output file
      File output = new File (console.nextLine());
      return output;
   }
   //method to count how many B's the person answered and stores them in an array
   public static int[] countB(String answers) { // takes the information from the user input that was stored in the answers variable
      int[] countB = new int[DIMENSIONS]; // method declares the array of 4, searching through 4 at a time
      for (int i = 0; i < answers.length(); i++) { // for every time i is less than the length of the answer continue on
         char t = answers.charAt(i); // declare a char and make it set to the character at place i in the answers
         if (t == 'B') { // if at that spot you find a B continue
            if (i % 7 == 0) { // found a B is the spot it is at evenly divisible by 7 aka in the 7th spot?
               countB[0]++; // ok then at it to the array in the first spot
            }
            if (i % 7 == 1 || i % 7 == 2) { //is the B found in
               countB[1]++;// ok then at it to the array in the second spot
            }
            if (i % 7 == 3 || i % 7 ==4) {

               countB[2]++;// ok then at it to the array in the third spot
            }
            if (i % 7 == 5 || i % 7 == 6) {
               countB[3]++; // ok then at it to the array in the fourth spot
            }

         }
      }
      return countB;
   }
   // method counts the A's in each dimension and return the total as an array
   public static int[] countA(String answers) {
      int[] countA = new int[DIMENSIONS];
      for (int i = 0; i < answers.length(); i++) {
         char t = answers.charAt(i);
         if (t == 'A') {
            if (i % 7 == 0) {
               countA[0]++;
            }
            if (i % 7 == 1 || i % 7 == 2) {
               countA[1]++;
            }
            if (i % 7 == 3 || i % 7 ==4) {
               countA[2]++;
            }
            if (i % 7 == 5 || i % 7 == 6) {
               countA[3]++;
            }
         }
      }
      return countA;
   }
我的程序在BABAAAABAAAAAAABAAAABBAAAAAAAAAAAA AAA中读取的字符串的示例，我目前有两个int数组int [] countA = new int [4]和int [] ...

Answer 1

要利用相似的代码，您可以尝试区分不同的部分并将它们用作参数。在您的情况下，数组countA和countB相似，唯一的文字'A'和'B'发生了变化。该原理称为'Encapsulate what varies'。

所以新方法可能看起来像这样：