如何打印出現次數

Question

我必須使用遞歸編寫程序，該程序會提示輸入一個單詞，然後計算字母a在該單詞中出現的次數。

我需要鍵入而不是+ count +才能使其工作？

這是我到目前爲止;

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

在你的代碼底部卸下此：

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

而且在main方法添加以下行：

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

Answer 2

你應該CAL的方法計數並顯示結果在主要方法。

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

你總是要遞歸，它只是你是否增加與否。你應該用代碼來表達它。

只用一隻（可讀）線替換整個方法：

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

打印結果的屏幕應該是從你的主要方法。

Answer 4

看來u必須不叫烏爾計數方法..

試試這個..

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

另一種方式可以

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

有在你的源犯了很多錯誤碼。您應該在短期內嘗試此操作，並閱讀Deitel的長期Java課程手冊，以提高您的編程技能。

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 
    } 
    } 
}