（Java）在句子中計算字母？

Question

以下是我的代碼：

char[] array = new char[26] ;

    int index = 0 ;
    int letter = 0 ;
    int countA = 0 ;

    String sentence = "Once upon a time..." ;

    if(sentence.contains("."))
    {
        String sentenceNoSpace = sentence.replace(" ", "").toLowerCase() ;
        String sentenceFinal = sentenceNoSpace.substring(0, sentenceNoSpace.indexOf(".")) ;
        char[] count = new char[sentenceFinal.length()] ;
        for (char c = 'a'; c <= 'z'; c++) 
        {
            array[index++] = c ;
            for(int i = 0; i < sentenceFinal.length(); i++)
            {
                if(sentenceFinal.charAt(i) == c)
                    count[letter++] = c ; 
                //if(sentenceFinal.charAt(i) == 'a')    
                    //countA++ ;   
            }

        }
        String result = new String(count) ; // Convert to a string.
        System.out.println("\n" + result) ;

        System.out.println("\nTotal number of letters is " + result.length()) ;
        System.out.println(countA) ;
    }
    else
    {
       System.out.println("You forgot a period. Try again.") ;
    }

我無法計算給定句子中有多少個a，b，c等。 我有一種方法可以做到，這就是這一部分

//if(sentenceFinal.charAt(i) == 'a')    
                //countA++ ;

我可以一直創建到z 有沒有更有效的方法？

注意：請勿使用Hashmap或任何其他高級技術。

Answer 1

無需消除空間。 這只是您正在做的其他工作。

int countOfLetters = 0 ;
String sentence = "Once upon a time..." ;
sentence = sentence.toLowerCase();
int[] countOfAlphabets = new int[26];
for (int i = 0; i < sentence.length(); i++) {
    if (sentence.charAt(i) >= 'a' && sentence.charAt(i) <= 'z') {
        countOfAlphabets[sentence.charAt(i) - 97]++;
        countOfLetters++;
    }
}

因此， countOfLetters將為您提供字母的總數。 例如，如果您要進行個體計數，則您想計數'c' ，

您可以通過訪問countOfAlphabets數組（例如countOfAlphabets['c' - 97] （97是'a'的ASCII值））來獲取它

Answer 2

使用int數組letterCounts將存儲每個字母的計數。 假設字母的大小寫可以忽略，則letterCounts數組的長度為26。

遍歷字符串的字符並更新數組中的相應整數。 使用其ASCII值查找相應的索引，如下所示。

letterCounts[c - 97]++

ASCII值“ a”為97，其計數需要存儲在索引0中。

這樣，從字符的ASCII值中減去97將得到該字符的相應索引。

注意：這是假設您要存儲小寫字母的計數。

Answer 3

不用地圖就可以了，但這會計算字符串中的所有字符。

您可能需要修改以排除空格等內容。

public class Main {

    public static void main(String[] args) {
        String sentence = "Once upon a time...";

        // Create an array of size 256 ASCII_SIZE
        int count[] = new int[256];
        int length = sentence.length();

        // Initialize count array index
        for (int i = 0; i < length; i++)
            count[sentence.charAt(i)]++;

        // Create an array of given String size
        char chars[] = new char[sentence.length()];
        for (int i = 0; i < length; i++) {
            chars[i] = sentence.charAt(i);
            int find = 0;
            for (int j = 0; j <= i; j++) {

                // If any matches found
                if (sentence.charAt(i) == chars[j])
                    find++;
            }

            if (find == 1) {
               System.out.println("Occurrence of " + sentence.charAt(i) + " is:" + count[sentence.charAt(i)]);

            }
        }
    }
}

哪個輸出：

Occurrence of O is:1
Occurrence of n is:2
Occurrence of c is:1
Occurrence of e is:2
Occurrence of   is:3
Occurrence of u is:1
Occurrence of p is:1
Occurrence of o is:1
Occurrence of a is:1
Occurrence of t is:1
Occurrence of i is:1
Occurrence of m is:1
Occurrence of . is:3

Answer 4

檢查下面的代碼您可以具有26個長度的數組，索引將根據字母的存在而增加。

public void getResult(){

        int [] charCount = new int [26];
        int countA = 0 ;

        String sentence = "Once upon a time..." ;

        if(sentence.contains("."))
        {
            String sentenceNoSpace = sentence.replace(" ", "").toLowerCase() ;
            String sentenceFinal = sentenceNoSpace.substring(0, sentenceNoSpace.indexOf(".")) ;

            char[] sentenceCharArray = sentenceFinal.toCharArray();
            //char a = 97;
            for (int i = 0; i <sentenceCharArray.length ; i++) {
                int index = sentenceCharArray[i] - 97 ;
                if(index >= 0 && index <= 26) {
                    charCount[index] += 1;
                }
            }


            System.out.print("Result : ");

            for (int i = 0; i < charCount.length ; i++) {
                System.out.print(charCount [i]+" , ");
            }


            System.out.println("\nTotal number of letters is " + sentenceCharArray.length) ;
        }
        else
        {
            System.out.println("You forgot a period. Try again.") ;
        }
    }

Answer 5

由於美國字母表中有26個字母，因此您可以使用大小為26的int[]

int[] letterCount = new int[26];

保留每個字母的計數，其中索引0代表'a'，1代表'b'，依此類推...

當您遍歷句子時，請檢查您所使用的字符是否為字母Character.isLetter() ，然后遞增表示字母的數組中的元素。

letterCount[letter - 'a']++;

我們從字母中減去“ a”來給我們正確的索引。

代碼樣例

package stackoverflow;

public class Question {

    public static void main(String[] args) {
        String sentence = "The quick brown fox jumps over the lazy dog.";
        int[] letterCount = new int[26];
        if (sentence.contains(".")) {
            // toLowerCase() the sentence since we count upper and lowercase as the same
            for (char letter : sentence.toLowerCase().toCharArray()) {
                if (Character.isLetter(letter)) {
                    letterCount[letter - 'a']++;
                }
            }

            // Display the count of each letter that was found
            int sumOfLetters = 0;
            for (int i = 0; i < letterCount.length; i++) {
                int count = letterCount[i];
                if (count > 0) {
                    System.out.println((char)(i + 'a') + " occurs " + count + " times");
                    sumOfLetters += count;
                }
            }

            System.out.println("Total number of letters is " + sumOfLetters);
        } else {
            System.out.println("You forgot a period.  Try again.");
        }
    }
}

結果

a occurs 1 times
b occurs 1 times
c occurs 1 times
d occurs 1 times
e occurs 3 times
f occurs 1 times
g occurs 1 times
h occurs 2 times
i occurs 1 times
j occurs 1 times
k occurs 1 times
l occurs 1 times
m occurs 1 times
n occurs 1 times
o occurs 4 times
p occurs 1 times
q occurs 1 times
r occurs 2 times
s occurs 1 times
t occurs 2 times
u occurs 2 times
v occurs 1 times
w occurs 1 times
x occurs 1 times
y occurs 1 times
z occurs 1 times
Total number of letters is 35

反駁問題

使用Java 8和使用String的chars()有什么問題？ 有了它，您可以用更少的代碼完成相同的事情。 對於字母總數，我們只使用String.replaceAll()並使用[^A-Za-z]模式從String刪除所有非字母，然后使用結果的length() 。

package stackoverflow;

import java.util.function.Function;
import java.util.stream.Collectors;

public class Question {

    public static void main(String[] args) {
        String sentence = "The quick brown fox jumps over the lazy dog.";

        System.out.println(sentence.toLowerCase().chars()
            // Change the IntStream to a stream of Characters
            .mapToObj(c -> (char)c)
            // Filter out non lower case letters
            .filter(c -> 'a' <= c && c <= 'z')
            // Collect up the letters and count them
            .collect(Collectors.groupingBy(Function.identity(), Collectors.counting())));

        System.out.println("Total letter count is " + sentence.replaceAll("[^A-Za-z]", "").length());               
    }
}

結果

{a=1, b=1, c=1, d=1, e=3, f=1, g=1, h=2, i=1, j=1, k=1, l=1, m=1, n=1, o=4, p=1, q=1, r=2, s=1, t=2, u=2, v=1, w=1, x=1, y=1, z=1}
Total letter count is 35

Answer 6

如果Regex不會被視為高科技If，則可以使用Regex解決。

想法很簡單： 刪除所有字母，然后從原始字符串長度中減去輸出以得到計數器

String sentence = "Once upon a time...";
String noLetterString = sentence.replaceAll("[a-zA-Z]", "");
int counterLetter = sentence.length() - noLetterString.length();
System.out.println("counter:" + counterLetter);

通過老派編程🙂這里的想法是相反的，僅添加字母

String sentence = "Once upon a time...";
String lowerCase = sentence.toLowerCase(); // to avoid comparison to UpperCase letters
StringBuilder counterStr = new StringBuilder();
for (char l : lowerCase.toCharArray()) {
    if (l >= 'a' && l <= 'z') {
        counterStr.append(l);
    }
}

System.out.println("counterStr:" + counterStr);
System.out.println("counter:" + counterStr.length());

Answer 7

這是更新代碼：

int[] array = new int[26] ;

    String sentence = "Once upon a time..." ;

    if(sentence.contains("."))
    {
        String sentenceNoSpace = sentence.replace(" ", "").toLowerCase() ;
        String sentenceFinal = sentenceNoSpace.substring(0, sentenceNoSpace.indexOf(".")) ;

        for (char c : sentenceFinal.toCharArray()) 
        {
            System.out.println(c+"  "+(c-97)); 
            array[c-97] += 1;     
        }

      // System.out.println("\n" + Arrays.toString(array)) ;

        for(int i=0; i< array.length;i++) {

            if(array[i] != 0) {

            char c = (char)(i+97);
            System.out.println(c+" occured "+ array[i]+" times");
          }
        }
    }
    else
    {
       System.out.println("You forgot a period. Try again.") ;
    }