不用递归解决汉诺塔问题

无聊的蛋疼。

#!/usr/bin/env python

def Hanoi(n):
    #递归解法
    def han(A,B,C,n):
        if n==1:
            print n,A,"-->",C
        else:
            han(A,C,B,n-1)
            print n,A,"-->",C
            han(B,A,C,n-1)
    han('A','B','C',n)
def Hanoi(n):
    #非递归解法
    even=('A','B','C')
    odd=('A','C','B')
    ls=[0 for i in range(n)]
    for i in range(1,2**n):
        deepth=n
        a=i
        while a&1==0:
            deepth-=1
            a=a>>1
        temp=odd if deepth%2==1 else even
        print n+1-deepth,temp[ls[deepth-1]%3],"-->",temp[(ls[deepth-1]+1)%3]
        ls[deepth-1]+=1

模拟递归

def Hanoi(n):
    a,b,c='A','B','C'
    status=1#1 开始 2左分支完毕 3右分支完毕
    stack=[]
    while True:
        #print a,b,c,status,n
        #raw_input()
        if n>1 and status==1:
            #左分支
            stack.append((a,b,c,status,n))
            n-=1
            a,b,c=a,c,b
        elif n==1:
            #底层
            print n,a,"-->",c
            a,b,c,status,n=stack.pop()
            if status==1 or status==2:
                status+=1
        elif n>1 and status==2:
            #右分支
            print n,a,"-->",c
            stack.append((a,b,c,status,n))
            n-=1
            a,b,c=b,a,c
            status=1
        else:
            #回退
            if len(stack)==0:
                return 
            a,b,c,status,n=stack.pop()
            if status==1 or status==2:
                status+=1