数据结构--汉诺塔--借助栈实现非递归---Java

 /*汉诺塔非递归实现--利用栈
  * 1.创建一个栈，栈中每个元素包含的信息：盘子编号，3个塔座的变量
  * 2.先进栈，在利用循环判断是否栈空，
  * 3.非空情况下，出栈，检查是否只有一个盘子--直接移动，否则就模拟前面递归的情况--非1的情况
  * 4.直到栈空就结束循环，就完成全部的移动。
  * */
 class Stack11{
     Towers[] tt = new Towers[20];
     int top = -1;

     public boolean isEmpty(){
         return top == -1;
     }

     public void push(Towers t){
         tt[++top] = t;
     }

     public Towers pop(){
         return tt[top--];
     }
 }

 class Towers{
     int diskN;
     char from;
     char inter;
     char to;
     public Towers(int diskN, char from, char inter, char to) {
         this.diskN = diskN;
         this.from = from;
         this.inter = inter;
         this.to = to;
     }

 }

 public class HannoTower_Stack {

     public static void main(String[] args) {
         Towers t1 = new Towers(3,'A','B','C');
         doTowers(t1);
     }

     private static void doTowers(Towers t1) {
         Stack11 stack = new Stack11();
         stack.push(t1);
         while(!stack.isEmpty()){
             Towers temp = stack.pop();
             //处理是一个盘子的情况--所有打印语句都从这里打印
             if(temp.diskN == 1){
                 System.out.println("Top disk " + "from " + temp.from + " to " + temp.to);
             }
             //注意处理移动的顺序本来是A-C-B,A-B-C,B-A-C.所以进栈的顺序相反
             else{
                 stack.push(new Towers(temp.diskN-1,temp.inter,temp.from,temp.to));
                 stack.push(new Towers(1,temp.from,temp.inter,temp.to));
                 stack.push(new Towers(temp.diskN-1,temp.from,temp.to,temp.inter));
             }
         }

     }

 }

执行结果：

Top disk from A to C
Top disk from A to B
Top disk from C to B
Top disk from A to C
Top disk from B to A
Top disk from B to C
Top disk from A to C