Question

对于我的问题，我有一个大于6+的计数列表。从该列表中我想制作一个列表，其中包含正好6张卡长的原始卡片的所有可能组合。（它们必须是唯一的，订单并不重要）

所以对象 01,02,03,04,05,06 和我一样 06,05,04,03,02,01

//STARTER list with more then 6 value's
List < ClassicCard > lowCardsToRemove = FrenchTarotUtil.checkCountLowCardForDiscardChien(handCards);

我找到并使用的解决方案：

public static List generateAllSubsetCombinations（object [] fullSet，ulong subsetSize）{ if（fullSet == null）{ 抛出新的ArgumentException（＆＃34; Value不能为null。＆＃34;，＆＃34; fullSet＆＃34;）; } else if（subsetSize＆lt; 1）{ 抛出新的ArgumentException（＆＃34;子集大小必须为1或更大。＆＃34;，＆＃34; subsetSize＆＃34;）; } else if（（ulong）fullSet.LongLength＆lt; subsetSize）{ 抛出新的ArgumentException（＆＃34;子集大小不能大于整个集合中的条目总数。＆＃34;，＆＃34; subsetSize＆＃34;）; }

    // All possible subsets will be stored here
    List<object[]> allSubsets = new List<object[]>();

    // Initialize current pick; will always be the leftmost consecutive x where x is subset size
    ulong[] currentPick = new ulong[subsetSize];
    for (ulong i = 0; i < subsetSize; i++) {
        currentPick[i] = i;
    }

    while (true) {
        // Add this subset's values to list of all subsets based on current pick
        object[] subset = new object[subsetSize];
        for (ulong i = 0; i < subsetSize; i++) {
            subset[i] = fullSet[currentPick[i]];
        }
        allSubsets.Add(subset);

        if (currentPick[0] + subsetSize >= (ulong)fullSet.LongLength) {
            // Last pick must have been the final 3; end of subset generation
            break;
        }

        // Update current pick for next subset
        ulong shiftAfter = (ulong)currentPick.LongLength - 1;
        bool loop;
        do {
            loop = false;

            // Move current picker right
            currentPick[shiftAfter]++;

            // If we've gotten to the end of the full set, move left one picker
            if (currentPick[shiftAfter] > (ulong)fullSet.LongLength - (subsetSize - shiftAfter)) {
                if (shiftAfter > 0) {
                    shiftAfter--;
                    loop = true;
                }
            }
            else {
                // Update pickers to be consecutive
                for (ulong i = shiftAfter+1; i < (ulong)currentPick.LongLength; i++) {
                    currentPick[i] = currentPick[i-1] + 1;
                }
            }
        } while (loop);
    }

    return allSubsets;
}

Answer 1

这个不是来自我，但它确实起作用了！

List <ClassicCard> lowCardsToRemove = FrenchTarotUtil.checkCountLowCardForDiscardChien(handCards);
var result = Combinator.Combinations(lowCardsToRemove, 6);

public static class Combinator
{
    public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int k)
    {
        return k == 0 ? new[] { new T[0] } :
          elements.SelectMany((e, i) =>
            elements.Skip(i + 1).Combinations(k - 1).Select(c => (new[] { e }).Concat(c)));
    }
}

如何从列表中获得所有独特的可能性？

1 个答案: