python - 使用递归查找数组的排列

Question

我想出了以下功能:

def permutations(elements, arr, index):
    if len(elements) == index:
        print arr
    for i in xrange(index, len(elements)):
        arr[index] = elements[i]
        permutations(elements, arr, index+1)

permutations([1,2,3], [0,0,0], 0)

但是，它正在打印:

[1, 2, 3]
[1, 3, 3]
[2, 2, 3]
[2, 3, 3]
[3, 2, 3]
[3, 3, 3]

我的递归有什么问题？

Answer 1

好吧，我发现这是一个 Java 字符串递归排列算法，我将其翻译成 Python。我认为这是最好的解决方案:

def permutation(aux, ls):
    if len(ls) == 0: 
        print(aux)
    else:
        for i in range(len(ls)):
            permutation(aux + [ls[i]], ls[0:i] + ls[i+1:len(ls)])

permutation([], [1, 2, 3])

这是输出:

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]