python - 非递归版本排列

Question

我有两个函数，prod 和 perm。它们非常相似。它们都使用递归。现在我不想用 for 循环替换递归。 prod2 工作正常，但 perm2 不正常，我该如何解决？

#Recursive version:

def prod(A,k):
    return [[]] if k==0 else [[a]+b for a in A for b in prod(A,k-1)]

def perm(A,k):
    return [[]] if k==0 else [[a]+b for a in A for b in perm([i for i in A if i!=a],k-1)]


#NonRecursive version:

def prod2(A,k):
    r=[[]]
    for i in range(k):
        r=[[a]+b for a in A for b in r]
    return r

def perm2(A,k):
    r=[[]]
    for i in range(k):
        r=[[a]+b for a in A for b in [i for i in r if i!=a ] ]
    return r

print prod([1,2,3],2)
print prod2([1,2,3],2)

print perm([1,2,3],2)
print perm2([1,2,3],2)

Answer 1

由于代码中的 r 变量包含列表，i != a 将始终为 True。以下是修复方法:

def perm2(A, k):
    r = [[]]
    for i in range(k):
        r = [[a] + b for a in A for b in [i for i in r if a not in i]]
    return r

或者简单地说:

def perm2(A, k):
    r = [[]]
    for i in range(k):
        r = [[a] + b for a in A for b in r if a not in b]
    return r