Question

我需要编写一个带有3个参数的河内递归算法。这就是我得到的：

#include <stdio.h> #include <stdio.h>

void hanoi(int m, int i, int j);
void move(int start, int end);

int main(){
    int n = 0;
    int i = 1;
    int j = 3;
    printf("Enter how many disks you want to move: ");
    scanf("%d", &n);
    int m = n;
    hanoi(m,i,j);

}

void hanoi(int m, int i, int j){    
    if (m==1)
    {
        move(i,j);
        return;
    }
    hanoi(m-1,i,2);
    move(i,j);
    hanoi(m-1,2,j);

}

void move(int start, int end){
    printf("A disk moves from position %d. to %d.\n", start,end);
}

n = 3的输出如下：

Enter how many disks you want to move: 3
A disk moves from position 1. to 2.
A disk moves from position 1. to 2.
A disk moves from position 2. to 2.
A disk moves from position 1. to 3.
A disk moves from position 2. to 2.
A disk moves from position 2. to 3.
A disk moves from position 2. to 3.

我查看了其他算法，我知道有很多算法用来解决这个古老的河内问题。但是，它们都有4个参数并使用字符，而我只想使用数字并在我的函数中省略辅助塔参数。我该如何解决？对于n = 1和n = 2，算法工作正常。

Answer 1

代码需要使用其他 peg，而不是@Karoly Horvath评论的void hanoi(int m, int i, int j){ if (m==1) { move(i,j); return; } int Other_peg = (1+2+3) - i - j; hanoi(m-1,i,Other_peg); move(i,j); hanoi(m-1,Other_peg,j); }

MyAdapter adapter = new MyAdapter(MyFragment fragmentA,...);

为什么我的河内算法塔不起作用？

1 个答案: