Question

我必须使用递归编写一个程序，它会提示输入一个单词，然后计算字母a出现在该单词中的次数。

我需要输入什么而不是+ count +才能使其正常工作？

这是我到目前为止所拥有的;

import java.util.Scanner;

public class test{
  public static void main(String[] args) {
    Scanner input=new Scanner(System.in);
    System.out.println("Enter String");
    String line = input.nextLine();
    char p = 'a';
  }  

  public static int count (String line, char p){
    int len = line.length();
    if ((len == 0) || (p == '\0')){   // case for empty string or nil char.
        return 0;                     // recursion ends here
    }
    String rest = line.substring(1);
    if (line.charAt(0) == p) {
        return count(rest, p) + 1;   // recurse on substring
    } else {

    } // end of if-else

    return count(rest, p);   // recurse on substring

  }

  System.out.println("Character a occurs " + count + " times ");

}

Answer 1

删除代码底部的内容：


{
   System.out.println("Character a occurs " + count + " times ");
}

在主要方法中添加以下行：

System.out.println("Character a occurs " + count(line,p) + " times ");

Answer 2

您应该调用方法计数并在main方法中显示结果。

public static void main(String[] args) {
    Scanner input=new Scanner(System.in);
    System.out.println("Enter String");
    String line = input.nextLine();
    char p = 'a';
    System.out.println("Character a occurs " + count(line, p) + " times ");
}

Answer 3

你总是需要递归，这只是你是否增加。你应该在代码中表达它。

您只需使用一条（可读）行替换整个方法：

public static int count(String line, char p) {
    return line.isEmpty() ? 0 : count(line.substring(1), p) + (line.charAt(0) == p ? 1 : 0);
}

将结果打印到屏幕应该来自您的主要方法。

Answer 4

好像你还没有调用你的计数方法..

试试这个..

public static void main(String[] args) {

   Scanner input=new Scanner(System.in);
    System.out.println("Enter String");
    String line = input.nextLine();
    char p = 'a';

    int count = count(line, p);

    System.out.println("Character a occurs " + count + " times ");

  }  
  public static int count (String line, char p)
  {
    int len = line.length();
    if ((len == 0) || (p == '\0'))   // case for empty string or nil char.
    return 0;                     // recursion ends here

    String rest = line.substring(1);
    if (line.charAt(0) == p)     
    {
      return count(rest, p) + 1;   // recurse on substring
    }
    else {

    } // end of if-else
    {
      return count(rest, p);   // recurse on substring
    }
  }

Answer 5

另一种方式可以是

import java.util.Scanner;
public class test{
   static int counter=0;
  public static void main(String[] args) {
    Scanner input=new Scanner(System.in);
    System.out.println("Enter String");
    String line = input.nextLine();
    char p = 'a';
    count(line, p);
    System.out.println("Character a occurs " + counter + " times ");
  }  

  public static void count (String line, char p){
    int len = line.length();
    if ((len == 0) || (p == '\0')){   // case for empty string or nil char.
        return;                     // recursion ends here
    }
    String rest = line.substring(1);
    if (line.charAt(0) == p) {
        counter++;   // recurse on substring
    } else {

    } // end of if-else
    count(rest, p);   // recurse on substring
  }
}

Answer 6

您的源代码中存在很多错误。您应该在短期内尝试这一点，并长期阅读Deitel＆amp; Deitel的Java How To Program一书，以提高您的编程技能。

import java.util.Scanner;

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

    System.out.println("Enter String");

    System.out.println("Character a occurs " + ((Integer)count(input.nextLine(), 'a')).toString() + " times."); 

    input.close();
  }  

  public static int count (String line, char p){
    if (line.length() == 0) {   // base case: no more chars to look at
        return 0;
    }

    if (line.charAt(0) == p) {
        return count(line.substring(1), p) + 1;   // recurse on substring
    }
    else {
        return count(line.substring(1), p);   // recurse on substring
    }
  }
}

如何打印出现次数

6 个答案: