Permutations全排列算法研究

来至于Codewars题目Permutations

在这个kata中，您必须创建输入字符串的所有排列，并删除重复项（如果存在）。这意味着，您必须以所有可能的顺序从输入中洗牌所有字母。
示例：
permutations（'a'）；#['a']
permutations（'ab'）；#['ab'，'ba']
permutations（'aabb'）；#['aabb'，'abab'，'abba'，'baab'，'baba'，'bbaa']
排列的顺序无关紧要。

我的解法

def permutations(nums):
    res = list(set(permute(nums)))
    return res

def permute(nums):
    res = []
    if len(nums) > 1:
        all_list = list(nums)
        for i in range(len(all_list)):
            heat = all_list.pop(i)
            res.extend([heat + one for one in permute(''.join(all_list))])
            all_list.insert(i, heat)
    else:
        res.extend(nums)
    return res

网友的解法

import itertools

def permutations(string):
    return list("".join(p) for p in set(itertools.permutations(string)))

itertools.permutations工具源码

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = list(range(n))
    cycles = list(range(n, n-r, -1))
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

附赠itertools.combinations工具源码

def combinations(iterable, r):
    # combinations('ABCD', 2) --> AB AC AD BC BD CD
    # combinations(range(4), 3) --> 012 013 023 123
    pool = tuple(iterable)
    n = len(pool)
    if r > n:
        return
    indices = list(range(r))
    yield tuple(pool[i] for i in indices)
    while True:
        for i in reversed(range(r)):
            if indices[i] != i + n - r:
                break
        else:
            return
        indices[i] += 1
        for j in range(i+1, r):
            indices[j] = indices[j-1] + 1
        yield tuple(pool[i] for i in indices)

permutations和combinations都是得到一个迭代器。
combinations方法重点在组合，permutations方法重在排列。