Java:特定字符报告的数量

Question

我制作了一个程序来计算短语中每个字母的数量

我希望结果文本返回的是

Enter a phrase

"Cow goes to the market"

There are 1 C('s), 3 O('s), 1 W('s), 1 G('s), 3 E('s), 1 S('s), 3 T('s).. /* you get the gist */ in your phrase.

我希望将结果连接成一个字符串，而不仅仅是一个多行报告。

import java.util.*;

public class charCount 
{
  public static void main(String[] args)
  {
   Scanner keyboard = new Scanner (System.in);
    //Prompt for user entry
    System.out.println("Enter a phrase: ");
    String letterN = keyboard.nextLine();

    charCount(letterN);
  }


public static int charCount(String letterN)
{ //variable initializations 
   
    int a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z;
    a = b = c = d = e = f = g = h = i = j = k = l = m = n = o = p = q = r = s = t = u = v = w = x = y = z = 0;

    int total = 0;
    total += 1;

//for loop 
for(int countN = 0; countN < letterN.length(); countN++) {
    switch(letterN.charAt(countN)) {
     case 'a':
        a++;
        break;
     case 'b':
        b++;
        break;
     case 'c':
    c++;
    break;
     case 'd':
    d++;
    break;
     case 'e':
    e++;
    break;
     case 'f':
    f++;
    break;
     case 'g':
        g++;
        break;
     case 'h':
        h++;
        break;
     case 'i':
    i++;
    break;
     case 'j':
    j++;
    break;
     case 'k':
    k++;
    break;
     case 'l':
    l++;
    break;
     case 'm':
        m++;
        break;
     case 'n':
        n++;
        break;
     case 'o':
    o++;
    break;
     case 'p':
    p++;
    break;
     case 'q':
    q++;
    break;
     case 'r':
    r++;
    break;
     case 's':
    s++;
    break;
     case 't':
    t++;
    break;
     case 'u':
    u++;
    break;
     case 'v':
    v++;
    break;
     case 'w':
    w++;
    break;
     case 'x':
    x++;
    break;
     case 'y':
    y++;
    break;
     case 'z':
    z++;
    break;
      
    }
    
}

    //Console logs
    System.out.println("Below are the individual character counts for your string");
    System.out.println("A: " + a);
    System.out.println("B: " + b);
    System.out.println("C: " + c);
    System.out.println("D: " + d);
    System.out.println("E: " + e);
    System.out.println("F: " + f);
    System.out.println("G: " + g);
    System.out.println("H: " + h);
    System.out.println("I: " + i);
    System.out.println("J: " + j);
    System.out.println("K: " + k);
    System.out.println("L: " + l);
    System.out.println("M: " + m);
    System.out.println("N: " + n);
    System.out.println("O: " + o);
    System.out.println("P: " + p);
    System.out.println("Q: " + q);
    System.out.println("R: " + r);
    System.out.println("S: " + s);
    System.out.println("T: " + t);
    System.out.println("U: " + u);
    System.out.println("V: " + v);
    System.out.println("W: " + w); 
    System.out.println("X: " + x);
    System.out.println("Y: " + y);
    System.out.println("Z: " + z);
  
    return total;    
  }  
}

我的问题是，我当然可以连接一个 System.out.println();使用 int vars 但它只是打印字母数。我事先不知道人们会输入什么，所以我需要我的程序检测字母是否 = true，从而返返回告中的字符和数字。

我一直在寻找解决方案，但一无所获。我还不能全神贯注，坦率地说，我不知道如何表达这个问题。

提前谢谢大家

Answer 1

用一个Map来存储你遇到过的人物，以及他们见过的次数。使用 TreeMap 意味着当您遍历 map 时，条目将按自然顺序排序

import java.util.*;

public class CharCount {
    public static void main(String[] args) {
        Scanner keyboard = new Scanner(System.in);
        //Prompt for user entry
        System.out.println("Enter a phrase: ");
        String letterN = keyboard.nextLine();

        charCount(letterN);
    }


    public static void charCount(String letterN) {
        Map<Character, Long> charMap = new TreeMap<>();
        for (char aChar : letterN.toCharArray()) {
            if (charMap.containsKey(aChar)) {
                charMap.put(aChar, charMap.get(aChar) + 1);
            }
            else {
                charMap.put(aChar, 1L);
            }
            // this if/else can be written using the ternary operator as:
            // charMap.put(aChar, charMap.containsKey(aChar) ? charMap.get(aChar) + 1 : 1L);
        }
        for (Character character : charMap.keySet()) {
            System.out.println(character + ": " + charMap.get(character));
        }
    }
}