在不使用庫的情況下計算字符串中字母出現的次數

Question

我的教授說我不能使用圖書館和東西。 我的庫中有代碼：

String phrase = keyboard.nextLine(); //Input.
int addr = phrase.length();
Map<Character, Integer> numChars = new HashMap<Character, Integer>(Math.min(addr, 26)); //there are 26 character in our alphabet. It makes a map and maps our the characters.

for (int i = 0; i < addr; ++i)//this area reads the string then, tells how many times each character appeared. Loops for each chracter.
{
  char charAt = phrase.charAt(i);

  if (!numChars.containsKey(charAt))
  {
    numChars.put(charAt, 1);
  }
  else if (numChars.containsKey(charAt))
  {
    numChars.put(charAt, 0);
  }
  else
  {
    numChars.put(charAt, numChars.get(charAt) + 1);
  }
}
  System.out.println(phrase);//outputs phrase written by user.
  System.out.println(numChars);//outputs results of the code above

// this code reads which one appeared the most.
 int FreqChar = 0;
 char frequentCh = ' ';

 for (int f = 0; f < phrase.length(); f++)
 {
    char poop = phrase.charAt(f);
    int banana = 0;
    for (int j = phrase.indexOf(poop); j != -1; j = phrase.indexOf(poop, j + 1))
    {
        frequentCh++;
    }
    if (banana > FreqChar)
    {
        FreqChar = banana;*/

這是我到目前為止沒有庫的程序。 我需要幫助將其轉換為數組。

    import java.util.*;

  public class LetCount
  {

  public static final int NUMCHARS = 26; //26 chars in alphabet.

  // int addr(char ch) returns the equivalent integer address for the letter
  // given in ch, 'A' returns 1, 'Z' returns 26 and all other letters return
  // their corresponding position as well. felt this is important.

  public static int addr(char ch)
  {
    return (int) ch - (int) 'A' + 1;
  }
// Required method definitions for (1) analyzing each character in an input
// line to update the appropriate count; (2) determining most frequent letter; 
// (3) determining least frequent letter; and (4) printing final results
// should be defined here.

  public static void main(String[] args)
  {
    Scanner keyboard = new Scanner(System.in);  // for reading input

    int[] count = new int [NUMCHARS]; // count of characters

    String phrase = keyboard.nextLine(); //Input.
    int addr = phrase.length();

       for(char ch = 'A'; ch <= 'Z'; ch++)
        {

    }
  }

Answer 1

這比在評論中更容易放在這里，但這並不是真正的答案:)

您已經有了一個不錯的開始-不用遍歷字母並找到匹配項，而是每次遇到一個字母時都要遍歷字符串並增加字母計數器（是的，這聽起來也很奇怪）。

首先將字符串轉換為小寫或大寫字母，您只需要計算字母，而無需根據現有代碼將其大小寫。

Answer 2

如果您只需要計算給定String中的所有字母並將其存儲在數組中，請嘗試以下操作：

String str = phrase.toUpperCase();    
for(int i = 0; i < str.length(); i++) {
    char c = str.charAt(i);
    int charPositionInArray = c - 'A';
    if(charPositionInArray < 26){
        count[charPositionInArray] += 1;
    }
}

另外，數組的索引從0開始，因此我假設您希望將'A'的計數存儲在count [0]中，即數組中的第一個位置。

此外，此代碼對非字母的任何字符均不執行任何操作。