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