如何将项保存到特定数组索引的次数返回?

I have a class called LetterCounter. This class processes strings by turning the letters of a string into chars and adding them to an array called counts[] with 26 positions, with 0 corresponding to 'a' and 25 corresponding to 'z'. This is what the process() method looks like.

我有一门课叫“信得过”。这个类通过将字符串的字母转换成字符来处理字符串，并将它们添加到一个名为counts[]的数组中，该数组有26个位置，0对应于“a”，25对应于“z”。process()方法是这样的。

    public class LetterCounter
{
    private static final int[] counts = new int[26];

/**
 * Method converts letters in a String into index positions in an array.
 * 0 corresponds to a, 1 to b, 2 to c, etc.
 * 
 * @param someString A String of letters
 */
public void process(String someString)
{
    for (int i = 0; i < someString.length(); i++)
    {
        char myChar = Character.toLowerCase(someString.charAt(i));
        int index = (myChar - 'a');
        if ((index >= 0) && (index <= 25))
        {
            counts[index]++;
        }
    }
}

I have a method in this class that I call getMostCommon() which iterates over the array and determines which position has the most stored items. It looks like this:

我在这个类中有一个名为getMostCommon()的方法，它遍历数组并确定哪个位置的存储项最多。它看起来像这样:

/**
 * Method finds the letter which appears most often.
 */
public char getMostCommon()
{
    int max = 0;

    for(int i = 0; i < counts.length; i++)
    {
        if(counts[i] > max)
        {
            max = i;
        }
    }

    char c = Character.toLowerCase((char)(max + 'a'));
    return c;
}

Here is the test class. I expect 't' to be the most common letter, but the method returns 'o' as the most common letter.

这是测试类。我希望't'是最常见的字母，但方法返回'o'作为最常见的字母。

public class CounterDemo1
{
    public static void main(String[] args)
    {
    //Constructs new LetterCounter called cc
    LetterCounter cc = new LetterCounter();

    //The letters in these strings will be processed into array index positions.
    cc.process("Bojack hates the troops");
    cc.process("Peanut butter is one word");
    cc.process("Use a pretty font");

    //Demonstrates printHistogram()
    cc.printHistogram();
    //Demonstrates getCount()
    System.out.println("There are " + cc.getCount('b') + " b's.");
    //Demonstrates getToalLetters()
    System.out.println("There are " + cc.getTotalLetters() + " total letters.");
    //Demonstrates getMostCommon()
    System.out.println("The most common letter is: " + cc.getMostCommon());
}

}

1 个解决方案

#1

You need to store the value of the maximun character frequency so far: max and the current character index: i.

到目前为止，您需要存储最大字符频率的值:max和当前字符索引:i。

You are comparing max with the frequency of the character i:if(counts[i] > max) and then assigning the value of i to max: max = i;

你在比较max与字符i:if(count [i] > max)的频率，然后将i的值赋给max: max = i;

You probably want to do this:

你可能想这样做:

public char getMostCommon()
{
     int max = 0;
     int current_index = 0; // current character index
     for(int i = 0; i < counts.length; i++)
     {
          if(counts[i] > max)
          {
               max = counts[i]; // Getting the value of the frequency
               current_index = i; // And the value of the character index
          }
      }
      // Finally parsing the index
      char c = Character.toLowerCase((char)(current_index + 'a'));
      return c;
}

#1