如何生成列表的所有排列?例如:

permutations([])
[]

permutations([1])
[1]

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

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

当前回答

def permutations(head, tail=''):
    if len(head) == 0:
        print(tail)
    else:
        for i in range(len(head)):
            permutations(head[:i] + head[i+1:], tail + head[i])

称为:

permutations('abc')

其他回答

def permutation(word, first_char=None):
    if word == None or len(word) == 0: return []
    if len(word) == 1: return [word]

    result = []
    first_char = word[0]
    for sub_word in permutation(word[1:], first_char):
        result += insert(first_char, sub_word)
    return sorted(result)

def insert(ch, sub_word):
    arr = [ch + sub_word]
    for i in range(len(sub_word)):
        arr.append(sub_word[i:] + ch + sub_word[:i])
    return arr


assert permutation(None) == []
assert permutation('') == []
assert permutation('1')  == ['1']
assert permutation('12') == ['12', '21']

print permutation('abc')

输出:['abc','acb','bac','bca','cab','cba']

def permuteArray (arr):

    arraySize = len(arr)

    permutedList = []

    if arraySize == 1:
        return [arr]

    i = 0

    for item in arr:

        for elem in permuteArray(arr[:i] + arr[i + 1:]):
            permutedList.append([item] + elem)

        i = i + 1    

    return permutedList

我不打算在一个新的行中穷尽所有的可能性,以使它有点独特。

我看到在这些递归函数中进行了很多迭代,而不是纯粹的递归。。。

所以对于那些连一个循环都不能遵守的人来说,这里有一个粗略的、完全不必要的完全递归的解决方案

def all_insert(x, e, i=0):
    return [x[0:i]+[e]+x[i:]] + all_insert(x,e,i+1) if i<len(x)+1 else []

def for_each(X, e):
    return all_insert(X[0], e) + for_each(X[1:],e) if X else []

def permute(x):
    return [x] if len(x) < 2 else for_each( permute(x[1:]) , x[0])


perms = permute([1,2,3])

用递归求解,遍历元素,取第i个元素,然后问自己:“其余项目的排列是什么”,直到没有更多的元素。

我在这里解释了解决方案:https://www.youtube.com/watch?v=_7GE7psS2b4

class Solution:
    def permute(self,nums:List[int])->List[List[int]]:
        res=[]
        def dfs(nums,path):
            if len(nums)==0:
                res.append(path)
            for i in range(len(nums)):
                dfs(nums[:i]+nums[i+1:],path+[nums[i]])
        dfs(nums,[])
        return res

另一种方法(无libs)

def permutation(input):
    if len(input) == 1:
        return input if isinstance(input, list) else [input]

    result = []
    for i in range(len(input)):
        first = input[i]
        rest = input[:i] + input[i + 1:]
        rest_permutation = permutation(rest)
        for p in rest_permutation:
            result.append(first + p)
    return result

输入可以是字符串或列表

print(permutation('abcd'))
print(permutation(['a', 'b', 'c', 'd']))