在递归函数中用尾递归替换 for 循环

Question

我正在尝试使以下函数完全尾递归，例如把那个讨厌的 for 循环从那里拿走。原因是我试图轻松地将解决方案转换为涉及使用显式堆栈的迭代解决方案。请指教

def permutations(A):
    P = []
    P2 = []
    permutations_recursive(A, [], P)
    permutations_tail_recursive(A, [], P2, 0)
    print(P2)
    return P

def permutations_recursive(first, last, perms):
    if len(first) == 0:
        perms.append(last)
    else:
        for i in range(len(first)):
            permutations_recursive(
                first[:i] + first[i+1:],
                last + [first[i]],
                perms)

Answer 1

关闭迭代模拟：

def permutations(A):
    P = []
    permutationsI(A, P)
    print(P)

def permutationsI(A, perms):
   stack = [(A, [])]
    while len(stack):
        first, last = stack.pop()
        if len(first):
            for i in range(len(first)):
                stack.append((first[:i] + first[i+1:],last + [first[i]]))
        else:
            perms.append(last)

permutations([1,2,3])
>>[[3, 2, 1], [3, 1, 2], [2, 3, 1], [2, 1, 3], [1, 3, 2], [1, 2, 3]]

Answer 2

一个完全递归的函数应该是：

def permutations_comp_recursive(first, last, perms, i):
    if len(first) == 0:
        perms.append(last)
    elif i == len(first):
        pass
    else:
        permutations_comp_recursive(first, last, perms, i+1)
        if first:
                permutations_comp_recursive(
                first[:i]+first[i+1:],
                last + [first[i]],
                perms, 0)

为了获得良好的性能，我建议。

编辑 1：现在下面应该是尾递归的，使用列表理解。这在 python 中使用 workarount 进行尾递归（并且省略了最后 2 个参数 - 结果作为 return 值传递）：

import itertools as it
class Recurse(Exception):
    def __init__(self, *args, **kwargs):
        self.args = args
        self.kwargs = kwargs

def recurse(*args, **kwargs):
    raise Recurse(*args, **kwargs)

def tail_recursive(f):
    def decorated(*args, **kwargs):
        while True:
            try:
                return f(*args, **kwargs)
            except Recurse as r:
                args = r.args
                kwargs = r.kwargs
                continue
    return decorated

@tail_recursive
def permutations_tail_recursive(first, last, direct=False):
    if len(first) == 0 or not all(first):
        return last
    else:
        l = len(first[0]) if direct else len(first)
        if last:
            return recurse([fi[:i]+fi[i+1:] for fi, i in it.product(first, range(l))],
                [last[j] + first[j][i] for j, i in it.product(range(len(last)), range(l))], True)
        else:
            return recurse([first[:i]+first[i+1:] for i in range(l)],
                [first[i] for i in range(l)], True)

这没有优化并且使用了循环。我不确定这个和上面没有循环的代码是否可以组合 - 可能会再次研究它。 itertools.permutations 可用于此应用程序。

在递归函数中用尾递归替换 for 循环

Replacing a for loop with tail recursion in a recursive function

python

algorithm

recursion

computer-science

tail-recursion