Question

我有一个河内拼图程序塔正在接近完成。我现在的问题是试图让用户输入正常工作。

如果他们输入'v'或'V'，则会显示解决拼图的步骤（因此输出将是'将光盘从S移动到D'等等）。否则，如果用户没有输入'v'或'V'，则程序继续解决拼图，显示总动作但不显示步骤。

我遇到的问题是选项不像他们应该的那样工作。

现在唯一不对的是当用户输入'v'或'V'时，移动没有正确显示。输出：

Enter the min number of discs : 
2
Press 'v' or 'V' for a list of moves
v
 Move disc from needle S to A
 Total Moves : 3

如果用户键入“v”或“V”，如何显示移动，如果用户键入除此之外的其他输出，则输出只显示“总移动”？ < / p>

这是我的代码：

 import java.util.*;

import java.util.Scanner;

public class TowerOfHanoi4 {
   static int moves=0;
   public static void main(String[] args) {

   System.out.println("Enter the min number of discs : ");
        Scanner scanner = new Scanner(System.in);
        int iHtMn = scanner.nextInt();       //iHeightMin         

        char source='S', auxiliary='D', destination='A';       //name poles or 'Needles'

   System.out.println("Press 'v' or 'V' for a list of moves");     
        Scanner show = new Scanner(System.in);
        String c = show.next();
        // char lstep='v', lsteps='V';       // grab option v or V

   if (c.equalsIgnoreCase("v")){        //if option is not v or V, execute code and only display total moves
    hanoi(iHtMn, source, destination, auxiliary); //else, user typed v or V and moves are displayed
    System.out.println(" Move disc from needle "+source+" to "+destination);
    System.out.println(" Total Moves : "+moves);
   } else {
      hanoi(iHtMn, source, destination, auxiliary);
      System.out.println(" Total Moves : "+moves);
     }  
   }


    static void hanoi(int htmn,char  source,char  destination,char  auxiliary)
      {
      if (htmn >=1)
          {
             hanoi(htmn-1, source, auxiliary, destination); // move n-1 disks from source to auxilary
              // System.out.println(" Move disc from needle "+source+" to "+destination); // move nth disk to destination
             moves++;     
             hanoi(htmn-1, auxiliary, destination, source);//move n-1 disks from auxiliary to Destination
          }
          // else (
      }
}

Answer 1

检查此行

if (c != lstep || c != lsteps){////}

即使c是v或V，这也会起作用如果c ='v'则c！='V'，所以这个'if test'适用于v或V. 将其更改为

if (c != lstep && c != lsteps){

Answer 2

您的代码的问题是else if条件，当您键入v或V时，应用程序会转到第一个if和条件为true，因此您的{{ 1}}永远不会被执行。

一个else if条件就足够了，试试这个

if

顺便说一句，您的代码中不需要多个扫描程序，只有一个就足够了。

也无需定义两个//if option is not v or V, execute code and only display total moves //else, user typed v or V and moves are displayed if (c == lstep || c == lsteps){ hanoi(iHeight, source, destination, auxiliary); System.out.println(" Move disc from needle "+source+" to "+destination); System.out.println(" Total Moves : "+moves); } else { hanoi(iHeight, source, destination, auxiliary); System.out.println(" Total Moves : "+moves); }来与它们进行比较，您可以轻松地执行此操作：

char

更新根据您的上次修改，您可以定义静态String c = show.next(); //if option is not v or V, execute code and only display total moves //else, user typed v or V and moves are displayed if (c.equalsIgnoreCase("v")){ hanoi(iHeight, source, destination, auxiliary); System.out.println(" Move disc from needle "+source+" to "+destination); System.out.println(" Total Moves : "+moves); } else { hanoi(iHeight, source, destination, auxiliary); System.out.println(" Total Moves : "+moves); }字段并将if条件移至boolean方法

检查pastebin here

试图让用户输入正常工作

2 个答案: