Question

我正在研究这个函数，在函数中我希望计算函数本身的迭代次数/递归次数。任何和所有帮助都会有所帮助！谢谢！

def runGenerations( L ):
    """ runGenerations keeps running evolve...
    """

    count = 0
    show(L)
    print(L)
    time.sleep(0.05)  

    if allOnes(L) == True:

        return L


    else:

        newL = evolve( L ) 
        return runGenerations( newL ) + 1

Answer 1

您可以在递归链上传递count个参数：

def runGenerations(L, count=1):
    show(L)
    print(L)

    if allOnes(L):
        print("result found after {} attempts".format(count))
        return L

     newL = evolve(L)
     return runGeneratons(newL, count+1) + 1

这个程序确实不需要递归。迭代解决方案将是：

def runGenerations_iterative(L):
    count = 1
    show(L)
    print(L)

    while not allOnes(L):
        L = evolve(L)
        count += 1
        show(L)
        print(L)

    print("result {} found after {} attempts".format(L, count))
    return L + count - 1

Answer 2

我想我明白了！

def runGenerations( L ):
    """ runGenerations keeps running evolve...
    """

    count = 0
    show(L)
    print(L)
    time.sleep(0.05)  

    if allOnes(L) == True:

        return 0


    else:

        newL = evolve( L ) 
        return 1 + runGenerations( newL )

计算函数内函数的递归次数

2 个答案: