简单易懂解决汉诺塔问题
作者:互联网
一、首先,汉诺塔问题说简单不简单,说容易不容易,所以就是难,,,,(孩子欠揍)
下面是我给大家提供的源代码,仅供参考。
import turtle class Stack: def __init__(self): self.items = [] def isEmpty(self): return len(self.items) == 0 def push(self, item): self.items.append(item) def pop(self): return self.items.pop() def peek(self): if not self.isEmpty(): return self.items[len(self.items) - 1] def size(self): return len(self.items) def drawpole_3():#画出汉诺塔的poles t = turtle.Turtle() t.hideturtle() def drawpole_1(k): t.up() t.pensize(10) t.speed(100) t.goto(400*(k-1), 100) t.down() t.goto(400*(k-1), -100) t.goto(400*(k-1)-20, -100) t.goto(400*(k-1)+20, -100) drawpole_1(0)#画出汉诺塔的poles[0] drawpole_1(1)#画出汉诺塔的poles[1] drawpole_1(2)#画出汉诺塔的poles[2] def creat_plates(n):#制造n个盘子 plates=[turtle.Turtle() for i in range(n)] for i in range(n): plates[i].up() plates[i].hideturtle() plates[i].shape("square") plates[i].shapesize(1,8-i) plates[i].goto(-400,-90+20*i) plates[i].showturtle() return plates def pole_stack():#制造poles的栈 poles=[Stack() for i in range(3)] return poles def moveDisk(plates,poles,fp,tp):#把poles[fp]顶端的盘子plates[mov]从poles[fp]移到poles[tp] mov=poles[fp].peek() plates[mov].goto((fp-1)*400,150) plates[mov].goto((tp-1)*400,150) l=poles[tp].size()#确定移动到底部的高度(恰好放在原来最上面的盘子上面) plates[mov].goto((tp-1)*400,-90+20*l) def moveTower(plates,poles,height,fromPole, toPole, withPole):#递归放盘子 if height >= 1: moveTower(plates,poles,height-1,fromPole,withPole,toPole) moveDisk(plates,poles,fromPole,toPole) poles[toPole].push(poles[fromPole].pop()) moveTower(plates,poles,height-1,withPole,toPole,fromPole) myscreen=turtle.Screen() drawpole_3() n=int(input("请输入汉诺塔的层数并回车:\n")) plates=creat_plates(n) poles=pole_stack() for i in range(n): poles[0].push(i) moveTower(plates,poles,n,0,2,1) myscreen.exitonclick()
把代码复制好,在python 3.7.2 Shell上运行,如图
可以看到效果,如图
这样汉诺塔问题就基本解决了,欢迎大家来评论-_- -_- -_-
标签:汉诺塔,400,self,简单,poles,易懂,plates,def 来源: https://www.cnblogs.com/liyanyinng/p/10611607.html