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我正在尝试在 Java 中编写一个 for 循环，该循环将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，我们还没有接触到数组或其他很多东西。（我意识到我两次声明了信，但此时我的大脑已经死了）这是我迄今为止尝试过的并且遇到了麻烦，感谢任何帮助：

好的，我根据建议更改了我的代码，但现在它只读取我句子的第一个单词？

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);

}
}
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10 回答 10

1

我看到了几个问题。首先，您有两个具有相同名称的变量。

其次，您的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），而不是仅仅假设您从第一个输入中获得 1 个字符，在第二个输入中获得 1 个句子。

于 2013-10-25T04:17:04.900 回答
0
  1. 您需要知道要搜索的字符。您可以使用 char charToSearch = letter.toCharArray()[0];
  2. 定义一个变量，比如count来计算给定字符串中某个字母的出现次数。
  3. 循环字符串并比较每个char，如果char等于要搜索的char，则count++；

示例--->

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);
于 2013-10-25T04:27:32.663 回答
0

希望这对您有所帮助。

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);
    }
}

样本输入输出：

Enter the string to search
how are you
Enter a character for which to search
o
No of Occurances : 2
于 2013-10-25T04:28:50.887 回答
0

你 if (sentence.length() <= 0) {的不对。改变你的条件，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));
于 2013-10-25T04:11:43.077 回答
0

尝试这个：

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);
于 2013-10-25T04:12:55.470 回答
0

尝试这个

忘记String letter = ""<-- 删除
忘记letter = in.next()<-- 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
于 2013-10-25T04:14:39.360 回答
0 
if (sentence.length() <= 0) {
            letter++;
}

您程序中的上述部分代码是错误的。除非您输入空字符串，否则这永远不会成立。

基本上这不是正确的逻辑。您将不得不使用直接比较。

于 2013-10-25T04:16:14.440 回答
0

无需循环：

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

输出：

numOfOccurences = 3
于 2013-10-25T04:19:09.520 回答
0

尝试 indexOf() 方法。它应该工作

于 2015-02-28T21:11:08.070 回答
0

阅读角色后，您的 Scanner 课程尚未移至下一行

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:)

于 2018-02-22T06:16:57.947 回答