解决河内拼图递归

-2

因此，我有一个程序，显示解决河内拼图的模拟。我的程序大部分工作正常，但不会在应该时终止。一旦所有光盘都在C钉（C C C）上，它应该终止，但是我的光盘仍然是第四张没有给出的光盘。应该只有三个光盘。任何帮助，将不胜感激！解决河内拼图递归

#include <iostream> 
using namespace std; 

const int num = 3; 
const char from_peg = 'A'; 
const char to_peg = 'B'; 
const char temp_peg = 'C'; 

char position[num]; 
void moveDiscs(int num,int disc,char source,char dest, char spare){ 
    if (disc == 0){ 
     position[disc]=dest; 
     cout<<"Moved disc "<<disc+1<<" to peg "<<dest; 
     cout<<" ("; 
     for(int i = 0;i<num;i++){ 
      cout<<position[i]<<" "; 
     } 
     cout<<")"<<endl; 
    }else{ 
     moveDiscs(num,disc-1,source,spare,dest); 
     position[disc]=dest; 
     cout<<"Moved disc "<<disc+1<<" to peg "<<dest; 
     cout<<" ("; 
       for(int i = 0;i<num;i++){ 
        cout<<position[i]<<" "; 
       } 
       cout<<")"<<endl; 
     moveDiscs(num,disc-1,spare,dest,source); 
    } 
} 
int main() { 

    cout<<"Starting Position for 3 discs are ("; 
    for(int i = 0;i<num;i++){ 
     position[i]='A'; 
     cout<<position[i]<<" "; 
    } 
    cout<<")"<<endl; 


    moveDiscs(3,3,from_peg,to_peg,temp_peg); 
    return 0; 
}

输出：

Starting Position for 3 discs are (A A A) 
Moved disc 1 to peg C (C A A) 
Moved disc 2 to peg B (C B A) 
Moved disc 1 to peg B (B B A) 
Moved disc 3 to peg C (B B C) 
Moved disc 1 to peg A (A B C) 
Moved disc 2 to peg C (A C C) 
Moved disc 1 to peg C (C C C) 
Moved disc 4 to peg B (C C C) 
Moved disc 1 to peg B (B C C) 
Moved disc 2 to peg A (B A C) 
Moved disc 1 to peg A (A A C) 
Moved disc 3 to peg B (A A B) 
Moved disc 1 to peg C (C A B) 
Moved disc 2 to peg B (C B B) 
Moved disc 1 to peg B (B B B)

来源

2017-08-02 Quinton Thompson

'position [disc] = dest;'使用'main'中第一个调用的else区块中的越界索引修改位置。 –

看起来像if块正在检查光盘0-3，而在else中它最终输出光盘+1。不会“给出”答案，但看看你如何比较光盘和输出应该让你走上正确的轨道。 – Das

[不是矢量乐趣？]（http://ideone.com/5MjTjb） - 问题的即时诊断。 – PaulMcKenzie

由于@Das指出光盘从0开始，但你传递的discs = 3数总盘从0到3实际上是4（同样的错误通常发生与数组长度）。您实际上应该设置discs等于2.

来源

2017-08-02 21:58:43

解决河内拼图递归

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