以間距遞歸地打印文字

Question

這是學校的作業。我無法理解我怎麼可以打印遞歸如下：

This was written by call number 2. 
This was written by call number 3. 
    This was written by call number 4. 
    This ALSO written by call number 4. 
    This ALSO written by call number 3. 
This ALSO written by call number 2. 
This ALSO written by call number 1. 

我不知道我是否應該是說明循環與遞歸或是否有遞歸打印所有的這一種方式。此外，我將如何去逆轉遞歸調用，以便從示例輸出的4開始？

這是我目前的輸出。

This was written by call number 2. 
This was written by call number 3. 
This was written by call number 4. 
This ALSO written by call number 1. 
    This ALSO written by call number 2. 
    This ALSO written by call number 3. 
    This ALSO written by call number 4. 

沒有在執行for循環B/C我不知道如果這部分也應該是遞歸的間距。

我的代碼：

public class Recursion { 

    public static void main(String[] args) { 
    for (int i = 2; i < 5; i++) { 
     System.out.println("This was written by call number " + i + "."); 
    } 
    recurse(4); 
    } 

    public static void recurse(int n) { 
    String temp = ""; 

    for (int i = 0; i < n; i++) { 
     temp += " "; 
    } 

    if (n < 2) { 
     System.out.println("This ALSO written by call number " + n + "."); 
    } 
    else { 
     recurse(n - 1); 
     System.out.println(temp + "This ALSO written by call number " + n + "."); 
    } 
} 

Answer 1

一個簡單的解決方案。

public static void main(String[] args) { 
    recurse(1); 
} 

public static void recurse (int n) { 
    if (n==5) return; 
    String temp=""; 
    for (int i=0;i<n;i++) temp += " "; 
    if (n!=1) { 
    System.out.println(temp + "This was written by call number " + n + "."); 
    } 
    recurse(n+1); 
    temp=" "; 
    for (int i=0;i<n;i++) temp += " "; 
    System.out.println(temp + "This ALSO was written by call number " + n + "."); 
} 

Answer 2

編寫最遞歸程序（特別是你給出的分配的）關鍵是要查找包含了同樣的問題的相似，但較小出現更大的問題。

在你的情況下，「更大的問題」是打印6行開始和結束「電話號碼2」。也就是說，電話號碼2到4的打印行。執行此操作的方法是：打印顯示「電話號碼2」的第一行，解決打印電話號碼3到4的4行4到的問題，並打印最後一行說「電話號碼2」。中間的部分是同一問題的發生率較小。這將是遞歸調用。

由於您的較大問題將以「電話號碼2」開頭，而您的較小問題將以較高的電話號碼開頭，因此我建議您安排一些事宜，以便撥打recurse(n+1)而不是recurse(n-1)。如果你這樣做，你需要第二個參數，以便知道何時停止遞歸 - 例如recurse(n+1, last)。

希望這將足以讓您在正確的軌道上思考。

Answer 3

試試這個：

public static void main(String[] args) { 
    recurse(1, true, 1); 
} 

public static void recurse(int n, boolean loop, int add) { 
    String temp = ""; 
    String out = ""; 

    for (int i = 0; i < n; i++) { 
     temp += " "; 
    } 

    if (add > 0) { 
     out = temp + "This was written by call number "; 
    } else { 
     out = temp + "This ALSO written by call number "; 
    } 

    if (n == 1 && !loop) { 
     System.out.println(out + n + "."); 
     return; 
    } else if (n == 1) { 
     recurse(n+add, false, add); 
    } else if (n == 5) { 
     add = add - 2 * add; 
     recurse(n+add, false, add); 
    } else { 
     System.out.println(out + n + "."); 
     recurse(n+add, false, add); 
    } 
} 

Answer 4

這是一個非常簡單的解決方案。另外請注意如何輕鬆獲取縮進字符串（通過子字符串）。遞歸過程非常簡單：打印數字，如果低於最大數字，則輸入數字較大的函數，然後返回。

class R{ 
     static final String spaces="         "; 
     public static void main(String[] args) { 
     rec3(1,4); 
     } 
     private static void rec3(int i, int max) { 
     if (i>1) System.out.printf("%sThis was written by call number: %d%n", spaces.substring(0, i-1), i); 
     if (i<max) rec3(i+1, max); 
     System.out.printf("%sThis was ALSO written by call number: %d%n", spaces.substring(0, i-1), i);  
     } 
    } 

Answer 5

感謝大家的幫助。我最終修改了@JoseLuis的解決方案。

public class Recursion { 

    public static void main(String[] args) { 
    recurse(1, 5); 
    } 

    public static void recurse(int n, int max) { 
    String temp = ""; 
    for (int i = 0; i < n; i++) { 
     temp += " "; 
    } 
    if (n == max) { 
     return; 
    } 
    if (n != 1) { 
     System.out.println(temp + "This was written by call number " + n + "."); 
    } 
    recurse(n + 1, max); 
    System.out.println(temp + "This ALSO was written by call number " + n + "."); 
    } 
}