Question

我正在尝试在Java中编写一个for循环来计算字符串中字母的出现次数。用户将输入要计数的字母和要搜索的字符串。这是一个非常基本的代码，我们尚未获得数组或其他许多代码。（我意识到我曾两次宣布信件，但此时我的大脑已经死了）这是我到目前为止所尝试的并且遇到麻烦，感谢任何帮助：

好的，我根据建议更改了我的代码，但现在它只是读了我的句子的第一个单词？

import java.util.Scanner;

public class CountCharacters {
public static void main(String[] args) {
    Scanner in = new Scanner(System.in);

    char letter;
    String sentence = "";
    System.out.println("Enter a character for which to search");
    letter = in.next().charAt(0);
    System.out.println("Enter the string to search");
    sentence = in.next();

    int count = 0;
    for (int i = 0; i < sentence.length(); i++) {
        char ch = sentence.charAt(i);
        if (ch == letter) {
            count++;
        }
    }
    System.out.printf("There are %d occurrences of %s in %s", count,
            letter, sentence);

}
}

Answer 1

我看到了几个问题。首先，您有两个具有相同名称的变量。

第二个if条件检查句子的长度是否大于0而不是检查字符是否相等。

Scanner in = new Scanner(System.in);

char inLetter = "";
String sentence = "";
System.out.println("Enter a character for which to search");
inLetter = in.next().charAt(0);
System.out.println("Enter the string to search");
sentence = in.next();

int letter = 0;
for (int i = 0; i < sentence.length(); i++) {
    char ch = sentence.charAt(i);
    if (inLetter == ch) {
        letter++;
    }
}

System.out.print(sentence.charAt(letter));

我还强烈建议验证输入（在上面的示例中没有完成），而不是假设你从第一个输入中得到1个字符，在第二个输入中得到1个句子。

Answer 2

您的if (sentence.length() <= 0) {不对。改变你的状况，如：

System.out.println("Enter a character for which to search");
letter = in.next();
System.out.println("Enter the string to search");
sentence = in.next();
char searchLet=letter.charAt(0); // Convert String to char
int letter = 0;
for (int i = 0; i < sentence.length(); i++) {
    char ch = sentence.charAt(i);
    if (searchLet== ch) {  // Check the occurrence of desired letter.
        letter++;
    }
}

System.out.print(sentence.charAt(letter));

Answer 3

试试这个：

Char letter = '';
String sentence = "";
System.out.println("Enter a character for which to search");
letter = in.next().charAt(0);
System.out.println("Enter the string to search");
sentence = in.next();

int count= 0;
for (int i = 0; i < sentence.length(); i++) {
    char ch = sentence.charAt(i);
    if (ch==letter) {
        count++;
    }
}
System.out.print(letter+" occurance:"+count);

Answer 4

试试这个

忘记String letter = ""＆lt; - 删除
忘记letter = in.next()＆lt; - 删除

    // There's no nextChar() method, so this is a work aroung
    char ch = in.findWithinHorizon(".", 0).charAt(0);

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

        if (sentence.charAt(i) == ch) {
            letter++;
        }
    }

    System.out.println(letter);  // print number of times letter appears

    // You don't want this
    System.out.print(sentence.charAt(letter)); // Makes no sense

Answer 5

if (sentence.length() <= 0) {
            letter++;
}

程序中的上述代码部分是错误的。除非你输入一个空字符串，否则这将永远不会成立。

基本上这不是正确的逻辑。您将不得不使用直接比较。

Answer 6

无需循环：

    String sentence = "abcabcabcd";
    String letter = "b";
    int numOfOccurences = sentence.length() - 
                          sentence.replaceAll(letter, "").length();
    System.out.println("numOfOccurences = "+numOfOccurences);

<强>输出：

numOfOccurences = 3

Answer 7

您需要知道要搜索的字符。您可以使用 char charToSearch = letter.toCharArray（）[0];
定义一个变量，例如 count ，以计算给定字符串中字母的出现次数。
循环字符串并比较每个char，如果char等于要搜索的char，则计算++;

实施例---＆GT;

int count = 0;
char charToSearch = letter.toCharArray()[0];
for (int i = 0; i < sentence.length(); i++) {
    if (sentence.charAt(i) ==  charToSearch) {
        count++;
    }
}

System.out.printf("Occurrences of a %s in %s is %d", letter, sentence, count);

Answer 8

希望这对你有所帮助。

import java.util.Scanner;

public class CountCharacters {
    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);

        System.out.println("Enter the string to search");
        String sentence = in.nextLine();

        System.out.println("Enter a character for which to search");
        String letter = in.next();

        int noOfOccurance = 0;

        for (int i = 0; i < sentence.length(); i++) {
            char dh=letter.charAt(0);
            char ch = sentence.charAt(i);
            if (dh==ch) {
               noOfOccurance++;
            }
       }
       System.out.print(noOfOccurance);
    }
}

示例输入输出：

Enter the string to search
how are you
Enter a character for which to search
o
No of Occurances : 2

Answer 9

尝试indexOf（）方法。它应该工作

Answer 10

您的Scanner类在读取字符

后没有移动到下一行

letter = in.next().charAt(0);

在读取输入字符串

之前添加另一个in.nextLine（）

    System.out.println("Enter a character for which to search");
    letter = in.next().charAt(0);
    in.nextLine();
    System.out.println("Enter the string to search");
    sentence = in.nextLine();

旧线程，但希望这会有所帮助：）

计算字符串中字母的出现次数

10 个答案: