记住经典的斐波拉契递归和阶乘递归转换为while规律

记住经典的斐波拉契递归和阶乘递归转换为while规律.它为实现更复杂转换提供了启发性思路.

# 斐波拉契--树形递归

def fab(n):
    if n<3:
        return n
    return fab(n-1)+fab(n-2)

def wfab(n):
    stacks=[(0,n,None)]
    while stacks:
        stg,n,value=stacks.pop()
        if stg==0:
            if n<3:
                res=n
            else:
                stacks.append((1,n,None))
                stacks.append((0,n-1,None))
        elif stg==1:
            stacks.append((2,n,res))
            stacks.append((0,n-2,None))
        else:
            res=value+res
    return res            

#阶乘--线性递归

def fac(n):
    if n<1:
        return 1
    return n*fac(n-1)

def wfac(n):
    stacks=[(0,n)]
    while stacks:
        stg,n=stacks.pop()
        if stg==0:
            if n<1:
                res=1
            else:
                stacks.append((1,n))
                stacks.append((0,n-1))
        elif stg==1:
            res=res*n
    return res

理解了原理,即每出现一次递归调用,先对当前栈做一个记号,再创建一个新栈,对应递归调用.比如,将fab略微改变一下,有更多系数,操作符和递归调用,次序也不是按照n-1,n-2这样来,也能比较容易地模拟出来:

def fab(n):
    if n<4:
        return n
    return (3*fab(n-3)-2*fab(n-1))*4*fab(n-2)

仿照上面可转化为:

def wfab(n):
    stacks=[(0,n,None)]
    while stacks:
        stg,n,value=stacks.pop()
        if stg==0:
            if n<4:
                res=n
            else:
                stacks.append((1,n,None))
                stacks.append((0,n-3,None))
        elif stg==1:
            stacks.append((2,n,3*res))
            stacks.append((0,n-1,None))
        elif stg==2:
            stacks.append((3,n,value-2*res))
            stacks.append((0,n-2,None))
        else:
            res=value*4*res
    return res

 再如:

arr=[1,5,10,25,50]

def xch(my,ars):
    if my==0:
        return 1
    if my<0 or not ars:
        return 0
    return xch(my,ars[1:])+xch(my-ars[0],ars)


def xch2(my,ars):
    stk=[(0,my,ars,None)]
    res = 0
    while stk:
        stg,my,ars,value = stk.pop()
        if stg==0:
            if my==0:
                res = 1
            elif my<0 or not ars:
                res = 0
            else:
                stk.append((1, my, ars, None))
                stk.append((0, my, ars[1:], None))
        elif stg==1:
            stk.append((2, my, ars, res))
            stk.append((0, my-ars[0], ars, None))
        else:
            res = res + value
    return res