如何使用python实现的汉诺塔的小程序

Posted 2020-12-13 Montai

python是面向对象的语言，但是做面向过程的操作，也是得心应手。

代码如下：

#!/usr/bin/env python
# -*- coding: utf-8 -*-
from requests.exceptions import RequestException

#定义栈，作为塔的数据结构
class Stack(object):
    def __init__(self, name):
        self.items = []
        self.name = name;

    def isEmpty(self):
        return self.items == []

    def push(self, item):
        self.items.append(item)

    def pop(self):
        return self.items.pop()

    def peek(self):
        return self.items[len(self.items) - 1]

    def size(self):
        return len(self.items)

#定义移动的块
class Block(object):
    def __init__(self, size, location):
        self.size = size
        self.location = location

    def setlocation(location):
        self.location = location

#移动汉诺塔
def moveblock(s1,s2):
    if(s1.size() > 0):
        if (s2.size() ==0) or ((s2.size() > 0) and (s2.peek().id > s1.peek().id)):
                b = s1.pop()
                s2.push(b)
                b.location = s2.name
                print (b.id, s1.name, "--->", s2.name)
        else:
            print("can not move")
#显示汉诺塔
def display(s1,s2,s3):
    disp(s1)
    disp(s2)
    disp(s3)

def disp(s):
    st=Stack(s.name)
    print(s.name)
    while(s.size()>0):
        b = s.pop()
        print(b.id)
        st.push(b)
    while(st.size()>0):
        s.push(st.pop())

#移动汉诺塔
def hanoi(n, a, b, c):
    if (n == 1):
        moveblock(a, c)
    else:
        hanoi(n - 1, a, c, b)
        moveblock(a, c)
        hanoi(n - 1, b, a, c)

#主程序，传入汉诺塔的层高
def main(n):
    #定义三个塔
    sLeft = Stack("left")
    sMiddles = Stack("middle")
    sRights = Stack("right")

    # 定义n层Block
    for i in range(n):
        block = Block(n-i,"left")
        sLeft.push(block)

    display(sLeft, sMiddles, sRights)
    hanoi(n, sLeft, sMiddles, sRights)
    display(sLeft,sMiddles, sRights)

if __name__ == ‘__main__‘:
    main(5)

程序运行结果：

left
1
2
3
4
5
middle
right

1 left ---> right
2 left ---> middle
1 right ---> middle
3 left ---> right
1 middle ---> left
2 middle ---> right
1 left ---> right
4 left ---> middle
1 right ---> middle
2 right ---> left
1 middle ---> left
3 right ---> middle
1 left ---> right
2 left ---> middle
1 right ---> middle
5 left ---> right
1 middle ---> left
2 middle ---> right
1 left ---> right
3 middle ---> left
1 right ---> middle
2 right ---> left
1 middle ---> left
4 middle ---> right
1 left ---> right
2 left ---> middle
1 right ---> middle
3 left ---> right
1 middle ---> left
2 middle ---> right
1 left ---> right

left
middle
right
1
2
3
4
5