/* Non-recursive solution to Towers of Hanoi */

汉诺塔非递归算法

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本文介绍了一种实现汉诺塔问题的非递归算法，并提供了完整的C语言代码示例。该算法通过循环而非递归的方式实现了汉诺塔的所有步骤，适用于希望深入理解汉诺塔算法逻辑及其实现细节的读者。

/* Non-recursive solution to Towers of Hanoi */

void main();

#include <stdio.h>

#define width (rings+1)

void main()
{
   int rings, last, next, x, z[500], s[3];

   printf("how many rings? "); scanf("%d",&rings);

   for(x=1; x<=rings; x++) /* put rings on first peg */
z[x]=width-x;
for(x=0; x<=2*width; x+=width) /* set base for each peg */
z[x]=1000;

/* if even number of rings, put first ring on second peg; if odd, on third */

if(rings%2==0)
{
last=1; s[2]=0; s[1]=1;
z[width+1]=z[rings];
}
else
{
last=2; s[1]=0; s[2]=1;
z[2*width+1]=z[rings];
}

printf("from 1 to %d/n",last+1); s[0]=rings-1;

while(s[0]+s[1]) /* while first and second pegs aren't empty */
{
/* next ring to move is smaller of rings on the two pegs not moved onto last */

if(last==0) next=z[width+s[1]]<z[2*width+s[2]]?1:2;
if(last==1) next=z[s[0]]<z[2*width+s[2]]?0:2;
if(last==2) next=z[s[0]]<z[width+s[1]]?0:1;

/* top ring of 'to' peg must be larger and an even 'distance' away */

if((z[next*width+s[next]]>z[last*width+s[last]])||
((z[last*width+s[last]]-z[next*width+s[next]])%2==0)) last=3-next-last;

      printf("from %d to %d/n",next+1,last+1);

      s[next]=s[next]-1; s[last]=s[last]+1; /* move from 'next' to 'last' peg */
      z[last*width+s[last]]=z[next*width+s[next]+1];
      }
}