1栈和队列--用栈来求解汉诺塔问题

最新推荐文章于 2021-11-10 12:29:05 发布

原创最新推荐文章于 2021-11-10 12:29:05 发布 · 784 阅读

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文章标签：

#栈 #递归 #游戏

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本文详细介绍了汉诺塔问题的两种解决方法：递归方法和非递归方法，并提供了具体的Java代码实现。对于递归方法，文章解释了如何通过递归函数实现汉诺塔的移动过程；对于非递归方法，则利用栈来模拟移动过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

修改游戏规则，只能移动一步，不能直接从最左侧到最右侧，必须经过中间过度。
当有N层塔时，打印最优移动过程和最优总步数。

递归方法
非递归方法，用栈来模拟三个塔

递归方法
（1）终止条件：移动最小号
（2）非最左到最右：三个步骤
（3）最左到最右：五个步骤

public static void main(String[] args) {
        int num = 4;

        // solution 1
        int steps1 = hanoiProblem1(num, "left", "mid", "right");
        System.out.println("It will move " + steps1 + " steps.");
        }


public static int hanoiProblem1(int num, String left, String mid,
            String right) {
        if (num < 1) {
            return 0;
        }
        return process(num, left, mid, right, left, right);
    }

    public static int process(int num, String left, String mid, String right,String from, String to) {
        if (num == 1) {
            if (from.equals(mid) || to.equals(mid)) {
                System.out.println("Move 1 from " + from + " to " + to);
                return 1;
            } else {
                System.out.println("Move 1 from " + from + " to " + mid);
                System.out.println("Move 1 from " + mid + " to " + to);
                return 2;
            }
        }
        if (from.equals(mid) || to.equals(mid)) {
            String another = (from.equals(left) || to.equals(left)) ? right : left;
            int part1 = process(num - 1, left, mid, right, from, another);
            int part2 = 1;
            System.out.println("Move " + num + " from " + from + " to " + to);
            int part3 = process(num - 1, left, mid, right, another, to);
            return part1 + part2 + part3;
        } else {
            int part1 = process(num - 1, left, mid, right, from, to);
            int part2 = 1;
            System.out.println("Move " + num + " from " + from + " to " + mid);
            int part3 = process(num - 1, left, mid, right, to, from);
            int part4 = 1;
            System.out.println("Move " + num + " from " + mid + " to " + to);
            int part5 = process(num - 1, left, mid, right, from, to);
            return part1 + part2 + part3 + part4 + part5;
        }
    }

2.非递归，用栈来模拟塔