汉诺塔问题
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题目描述
In the classic problem of the Towers of Hanoi, you have 3 towers and N disks of different sizes which can slide onto any tower. The puzzle starts with disks sorted in ascending order of size from top to bottom (i.e., each disk sits on top of an even larger one). You have the following constraints:
(1) Only one disk can be moved at a time.
(2) A disk is slid off the top of one tower onto another tower.
(3) A disk cannot be placed on top of a smaller disk.
Write a program to move the disks from the first tower to the last using stacks.
Example1:
Input: A = [2, 1, 0], B = [], C = [] Output: C = [2, 1, 0]
Example2:
Input: A = [1, 0], B = [], C = [] Output: C = [1, 0]
Note:
A.length <= 14
代码结果
运行时间: 16 ms, 内存: 16.1 MB
/*
* 思路:
* 1. 使用 Java Stream API 是不太可能直接实现递归逻辑,但可以使用 Stream 处理一些数据。
* 2. 这里的代码更适合用传统方式实现递归来处理汉诺塔问题。
*/
import java.util.Stack;
import java.util.stream.Stream;
public class HanoiTowerStream {
public void hanoi(int n, Stack<Integer> from, Stack<Integer> to, Stack<Integer> aux) {
if (n == 0) return;
hanoi(n - 1, from, aux, to);
to.push(from.pop());
hanoi(n - 1, aux, to, from);
}
public static void main(String[] args) {
Stack<Integer> A = new Stack<>();
Stack<Integer> B = new Stack<>();
Stack<Integer> C = new Stack<>();
Stream.of(2, 1, 0).forEach(A::push);
new HanoiTowerStream().hanoi(A.size(), A, C, B);
System.out.println(C);
}
}
解释
方法:
本题解采用了经典的递归方法来解决汉诺塔问题。递归方法分为几个步骤:1. 将顶部的n-1个盘子从源柱子A使用辅助柱子C移动到目标柱子B。2. 将最大的盘子直接从源柱子A移动到目标柱子C。3. 最后,将那n-1个盘子从B通过A移动到C。递归的基础情况是当只有一个盘子时,直接将其从A移动到C。
时间复杂度:
O(2^n)
空间复杂度:
O(n)
代码细节讲解
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在递归解决汉诺塔问题中,为何选择将n-1个盘子先移动到辅助柱子B而不是直接到目标柱子C?
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递归函数中,`self.hanota(A[1:], C, B)`这一行代码的具体含义是什么?如何理解此处的递归调用?
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特别是对于边界情况,当A中只有一个盘子时,解决方案是什么样的?这是否考虑了所有可能的边界情况?
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