Question

我的教授说我不能使用图书馆和东西。我有我的代码库：

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;*/

到目前为止，我的程序没有库。我需要帮助将其转换为数组。

    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

这比评论更容易放在这里，但它并不是真正的答案：）

你有一个很好的开始 - 而不是通过字母和找到匹配，每次遇到一个字母时，通过字符串并递增你的字母计数器（是的，这听起来也很奇怪）。

首先将字符串转换为小写或大写字母，您只需根据现有代码计算字母数，而不是它是低或高。

Answer 2

如果您只需计算给定String中的所有字母并将其存储在数组中，请尝试以下操作：

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;
    }
}

此外，数组的索引从0开始，所以我假设您希望将'A'的计数存储在count [0]中，即数组中的第一个位置。

此外，此代码不会对任何非字母表的字符执行任何操作。

计算字母在不使用库的情况下出现在字符串中的次数

2 个答案: