Question

我有一个这样的列表：

from itertools import permutations
l = [1,1,1,1,1,1,1,2]

原始列表中重复的'1'条目意味着不同的排列仅取决于'2'在输出中出现的位置；因此只有8个不同的排列。但是permutations（）函数将生成所有阶乘（8）= 40320排列。我知道我可以在事发后使用set（）函数删除重复的输出，但是算法仍然是O（N！），我想要更有效的方法。

Answer 1

这里有几种不使用set的有效解决方案，基本上是关于避免插入重复的元素。

# To handle duplication, just avoid inserting a number before any of its duplicates.
def permuteUnique1(nums):
    ans = [[]]
    for n in nums:
        new_ans = []
        for l in ans:
            for i in range(len(l) + 1):
                new_ans.append(l[:i] + [n] + l[i:])
                if i < len(l) and l[i] == n: break  # handles duplication
        ans = new_ans
    return ans


# Duplication happens when we insert the duplicated element before and after the same element,
# to eliminate duplicates, just insert only after the same element.
def permuteUnique2(nums):
    if not nums:
        return []
    nums.sort()
    ans = [[]]
    for n in nums:
        new_ans = []
        l = len(ans[-1])
        for seq in ans:
            for i in range(l, -1, -1):
                if i < l and seq[i] == n:
                    break
                new_ans.append(seq[:i] + [n] + seq[i:])
        ans = new_ans
    return ans


# Build the list of permutations one number at a time, insert the number into each already built permutation
# but only before other instances of the same number, never after.
def permuteUnique3(nums):
    perms = [[]]
    for n in nums:
        perms = [p[:i] + [n] + p[i:]
                 for p in perms
                 for i in range((p + [n]).index(n) + 1)]
    return perms


# or as "one-liner" using reduce:
from functools import reduce
def permuteUnique4(nums):
    return reduce(lambda perms, n: [p[:i] + [n] + p[i:]
                                    for p in perms
                                    for i in range((p + [n]).index(n) + 1)],
                  nums, [[]])

您可以在LeetCode中找到更多的解决方案和说明。希望对您有所帮助，如有其他问题，请发表评论。：）

python中具有重复的置换（不要使用set（）或uniform（）方法）

1 个答案: