给定词典索引的多集置换算法

Question

我试图找到一个有效的算法来寻找一个多集的排列，给定一个索引。在

例如：给定{1, 3, 3}。所有按字典序升序排列的排列都是{133, 313, 331}。这些元素被索引为{0, 1, 2}。给定index=2，结果是331。在

I found an algorithm查找给定词典索引的集合的置换。他的算法是有效的：O（n^2）。在

但是，该算法是在一个适当的集合上测试的（例如{1, 2, 3}），而在我的测试中并不正确。我在这里描述他的python代码，这样您就可以很容易地理解了。在

from math import factorial, floor #// python library
from math import factorial, floor #// python library
i=5 #// i is the lexicographic index (counting starts from 0)
n=3 #// n is the length of the permutation
p = range(1,n+1) #// p is a list from 1 to n
for k in range(1,n+1): #// k goes from 1 to n
    d = i//factorial(n-k) #// use integer division (like division+floor)
    print(p[d]),
    p.remove(p[d])   #//delete p[d] from p
    i = i % factorial(n-k) #// reduce i to its remainder

Answer 1

# Python 2
from collections import Counter
from math import factorial


def count_permutations(counter):
    values = counter.values()
    return (
        factorial(sum(values))/reduce(lambda a, v: a * factorial(v), values, 1)
    )


def permutation(l, index):
    l = sorted(l)

    if not index:
        return l

    counter = Counter(l)
    total_count = count_permutations(counter)
    acc = 0
    for i, v in enumerate(l):

        if i > 0 and v == l[i-1]:
            continue

        count = total_count * counter[v] / len(l)

        if acc + count > index:
            return [v] + permutation(l[:i] + l[i + 1:], index - acc)

        acc += count

    raise ValueError("Not enough permutations")

似乎像预期的那样有效

^{pr2}$

时间复杂性：O(n^2)。在