Chapter 3 Stacks and Queues - 3.4-优快云博客

本文链接：https://blog.youkuaiyun.com/cheng_tian/article/details/7709995

本文介绍了一个经典的递归问题——汉诺塔，并提供了一种使用堆栈实现的递归解决方案。该方案遵循汉诺塔游戏的基本规则，通过递归调用实现了盘子从初始杆到目标杆的移动。

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Problem 3.4: In the classic problem of the Towers of Hanoi, you have 3 rods and N disks fo different sizes which can slide onto any tower. The puzzle starts with disks sorted in ascending order of size from top to bottom (e.g., each disk sits on top of an even larger one). You have the following constrains:
(A) Only one disk can be moved at a time.
(B) A disk is slid off the top of one rod onto the next rod.
(C) A disk can only be placed on top of a larger disk.
Write a program to move the disks from the first rod to the last using Stacks.

So classic, so recursive:P

from stack import *
rods = [stack(), stack(), stack()]
def Hanoi(n, rod_from, rod_to):
    if n == 2:
        Move(rod_from, 3 - (rod_from + rod_to))
        Move(rod_from, rod_to)
        Move(3 - (rod_from + rod_to), rod_to)
    else:
        Hanoi(n - 1, rod_from, 3 - (rod_from + rod_to))
        Move(rod_from, rod_to)
        Hanoi(n - 1, 3 - (rod_from + rod_to), rod_to)
def Move(rod_from, rod_to):
    print "Move from", rod_from, "to", rod_to
    rods[rod_to].push(rods[rod_from].pop())

# Test case
if __name__ == "__main__":
    # range(1, 8)[::-1] gets a reverse list
    # Sequence slicing [starting-at-index : but-less-than-index [ : step]].
    # Start defaults to 0, end to len(sequence), step to 1
    for i in range(1, 8)[::-1]:
        rods[0].push(i)
    Hanoi(7, 0, 2)
    for i in range(0, 7):
        print "rods[2].pop():", rods[2].pop()