Recursive implementation of the Tower of Hanoi problem

I brought the Tower of Hanoi todayC语言实现代码.

汉诺塔（Tower of Hanoi）,又称河内塔,是一个源于印度古老传说的益智玩具.大梵天创造世界的时候做了三根金刚石柱子,在一根柱子上从下往上按照大小顺序摞着64片黄金圆盘.大梵天命令婆罗门把圆盘从下面开始按大小顺序重新摆放在另一根柱子上.并且规定,在小圆盘上不能放大圆盘,在三根柱子之间一次只能移动一个圆盘.

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There are three pillars as follows：

游戏规则如下：在a柱上会有n个盘子,And the big plate is below,The small plate is on top,每次只能移动一个盘子,Press the plate from largest to smallestcThe pillars are arranged upwards.

Suppose there is only one plate,Then we take the plate directly fromacolumn moved toc柱,即a->c.

假设aThere are two plates on the column,Then we should move the plate to the smaller oneb柱,Then move the larger platec盘,再将bThe plate on the column is moved toc柱,即 a->b , a->c , b->c This also completes the mission of the game.

假设a柱上有三个盘子,Then we should remove the smallest plate froma柱移到c柱,Then remove the smaller plate froma柱移到b柱,Then remove the smallest plate fromc柱移到b柱,再将最大的盘子从a柱移到c柱,再将bThe column with the smallest column moves toa柱,再将bMove the column to the smaller platec柱,最后把aMove the column plate toc柱.即 a->c , a->b , c->b , a->c , b->a , b->c , a->c .

下面是代码实现.

#include<stdio.h>
void move (char pos1,char pos2)
{
    
    printf(" %c->%c ",pos1,pos2);
}
void Hanoi (int n ,char pos1,char pos2,char pos3)
{
    
    if(n==1)
	{
    
	    move(pos1,pos3);
	}
	else
	{
    
	    Hanoi(n-1,pos1,pos3,pos2);
		move(pos1,pos3);
		Hanoi(n-1,pos2,pos1,pos3);
	}
}
int main ()
{
    
	 Hanoi(1,'a','b','c');
	 printf("\n");
	 Hanoi(2,'a','b','c');
	 printf("\n");
	 Hanoi(3,'a','b','c');
	 printf("\n");
}

运行结果如下：

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