数据结构--汉诺塔--借助栈实现非递归---Java

 1 /*汉诺塔非递归实现--利用栈
 2  * 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
 3  * 2.先进栈，在利用循环判断是否栈空，
 4  * 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
 5  * 4.直到栈空就结束循环，就完成全部的移动。
 6  * */
 7 class Stack11{
 8     Towers[] tt = new Towers[20];
 9     int top = -1;
10
11     public boolean isEmpty(){
12         return top == -1;
13     }
14
15     public void push(Towers t){
16         tt[++top] = t;
17     }
18
19     public Towers pop(){
20         return tt[top--];
21     }
22 }
23
24 class Towers{
25     int diskN;
26     char from;
27     char inter;
28     char to;
29     public Towers(int diskN, char from, char inter, char to) {
30         this.diskN = diskN;
31         this.from = from;
32         this.inter = inter;
33         this.to = to;
34     }
35
36 }
37
38 public class HannoTower_Stack {
39
40     public static void main(String[] args) {
41         Towers t1 = new Towers(3,‘A‘,‘B‘,‘C‘);
42         doTowers(t1);
43     }
44
45     private static void doTowers(Towers t1) {
46         Stack11 stack = new Stack11();
47         stack.push(t1);
48         while(!stack.isEmpty()){
49             Towers temp = stack.pop();
50             //处理是一个盘子的情况--所有打印语句都从这里打印
51             if(temp.diskN == 1){
52                 System.out.println("Top disk " + "from " + temp.from + " to " + temp.to);
53             }
54             //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
55             else{
56                 stack.push(new Towers(temp.diskN-1,temp.inter,temp.from,temp.to));
57                 stack.push(new Towers(1,temp.from,temp.inter,temp.to));
58                 stack.push(new Towers(temp.diskN-1,temp.from,temp.to,temp.inter));
59             }
60         }
61
62     }
63
64 }

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