汉诺塔问题python

count = 0
def hanoi(n,src,mid,dst):
    global count
    if n == 1:
        print("{}:{}->{}".format(1,src,dst))
        count += 1
    else:
        hanoi(n-1,src,dst,mid)
        print("{}:{}->{}".format(n, src, dst))#第n个圆盘从第src位置移动到dst位置
        count += 1
        hanoi(n-1,mid,src,dst)
hanoi(3,"A","B","C")
print(count)
#不显示对应序列号圆盘的简练写法

def move(n, a, b, c):
    if n == 1:
        print(a, ‘-->‘, c)
    else:
        move(n-1,a,c,b)
        print(a,‘-->‘,c)
        move(n-1,b,a,c)
move(3,"A","B","C")

汉诺塔问题python

原文：https://www.cnblogs.com/zhenshj/p/8830267.html