组合列表以获取项目组的最有效方法 <= 但尽可能接近长度 X

Question

假设我有一个列表列表，例如：

 [ [ a1, a2, a3, a4, a5], [b1, b2, b3, b4, b5, b6, b7, b8], [c1, c2, c3, c4, c5, c6], [d1, d2, d3, d4] ]

比较所有列表项的长度并将它们组合成一个长度尽可能接近但小于或等于 x 的列表的最简单方法是什么？

所以对于上面的例子，x=12：

[ [a1, a2, a3, a4, a5, c1, c2, c3, c4, c5, c6], [b1, b2, b3, b4, b5, b6, b7, b8, d1, d2, d3, d4] ]

输出中各个组（例如 a、b、c...）的顺序并不重要，但不能分解各个组。

我知道我可以，例如，获取第一组的长度，然后按顺序获取每个后续​​组的长度，如果它们的 sum = x 然后弹出这些列表并将它们的项目附加到返回的列表中，如果然后不再检查每个组的长度之和是否=x-1，如果是，则弹出并追加，然后继续使用长度之和=x-2等，直到输入列表为空。

对于像给出的示例这样的小团体来说，这会很好，但是当输入列表的 len 变得非常大时呢？有没有更高效的方法/算法？

Answer 1

这似乎是一个装箱问题。

我的greedy algorithm with swapping and pre-sorting 应该可以工作，但您需要调整子列表长度以适应图片高度，并根据您的 x 和项目的总长度计算初始 bin-count。

Answer 2

这个解决方案不一定找到长度最优的解决方案，因为它只考虑加入两个列表，但它在 O(n) 中运行。

#!/usr/bin/env python3

import sys
import random

def foo(xs, n):
    bins = {}
    mid = (n+1)//2

    # bin list positions by length
    for i, x in enumerate(xs):
        bucket = len(x)
        if bucket > n:
            raise RuntimeError("invalid input")

        bins.setdefault(bucket, []).append(i)

    # take out ones that are the desired length already
    out = [xs[x] for x in bins[n]] if n in bins else []
    bins.pop(n, None)

    # find complements for the upper half of buckets
    for i in list(bins.keys()):
        if i < mid:
            continue

        candidates = sorted([x for x in list(bins.keys()) if x <= n-i], reverse=True)

        while i in bins and bins[i]:
            x = bins[i].pop()

            for j in list(candidates):
                if j not in bins or not bins[j]:
                    candidates = candidates[1:]
                    continue

                y = bins[j].pop()
                if not bins[j]:
                    del bins[j]

                out.append(xs[x] + xs[y])
                break
            else:
                # complement not found
                out.append(xs[x])


    # add lists with no complements from the lower half
    out += [xs[y] for ys in bins.values() for y in ys]

    return out

_check_n = 0
def check(n, xs):
    ys = list(foo(xs, n))

    try:
        for y in ys:
            assert len(y) <= n
        print(".", end="")
        n+=1
        if n % 10 == 0:
            sys.stdout.flush()
    except:
        print("n, xs =", (n, xs))
        print("ys =", ys)
        raise

if __name__ == "__main__":
    n, xs = 12, [ ["a1", "a2", "a3", "a4", "a5"], ["b1", "b2", "b3", "b4", "b5", "b6", "b7", "b8"], ["c1", "c2", "c3", "c4", "c5", "c6"], ["d1", "d2", "d3", "d4"] ]
    check(n, xs)

    cases = 10**4
    max_n = 10**2
    max_input_len = 10**3

    for i in range(cases):
        n = random.randint(1, max_n)
        xs = [[1] * random.randint(1, n) for j in range(random.randint(1, max_input_len))]
        check(n, xs)