Question

我正在为.NETMF应用程序编写代码，需要创建一个接受2个参数int min, int max的函数，并从生成的2D Arraylist {{1}返回一个整数数组returnedArr } arr内的整数之和将在returnedArr和min之间，并且在调用方法之前我也不知道max的大小或内容是什么。

将选择arr内每个Arraylist只有1个整数。

arr

示例为public ArrayList GetNums(int min, int max) { //arr will be structured like this //var arr = new ArrayList //{ // new ArrayList {10, 34, 56, 60}, // new ArrayList {3, 23, 56, 78, 65, 42}, // new ArrayList {53, 56, 76} //}; var returnedArr = new ArrayList(); var arr = GenerateArraylistValues(); //solution code here... return returnedArr; } var returnedArr = new ArrayList(); var arr = GenerateArraylistValues(); //solution code here... return returnedArr; } 返回arraylist可以是GetNums(130,140);

另外请记住，我只限于使用Arraylists，因此我无法使用通用列表，我认为某种Arraylist{56,23,53}可以完成这项工作但不能让我了解这是如何工作的，因为我不知道foreach loop的大小将是什么样的

----------- SAMPLE＆＃39; arr＆＃39; VALUES ---------

arr

Answer 1

如果＆＃39; arr＆＃39;使用递归和蛮力，并不是很大：

    public ArrayList GetNums(int min, int max, ArrayList arr)
    {
        var results = new ArrayList();

        //nothing in the arr?
        if (arr.Count == 0)
            return results;

        //arr has only one list inside
        if(arr.Count == 1)
        {
            foreach(int a in (ArrayList)arr[0])
            {
                if(a >= min && a <= max)
                {
                    var r = new ArrayList();
                    r.Add(a);
                    results.Add(r);
                }
            }
            return results;
        }

        //arr has two or more lists inside
        ArrayList firstList = (ArrayList)arr[0];
        ArrayList remainingArr = new ArrayList();
        for(int i = 1; i < arr.Count; i++)
        {
            remainingArr.Add(arr[i]);
        }

        foreach (int a in firstList)
        {
            var tempResults = GetNums(min - a, max - a, remainingArr);
            foreach(ArrayList result in tempResults)
            {
                var newResult = new ArrayList();
                newResult.Add(a);
                newResult.AddRange(result);
                results.Add(newResult);
            }
        }
        return results;
    }

结果是一系列解决方案。您的示例arr的第一个有效解决方案是{10,65,56}

Answer 2

如果可以接受每次都有相同的最小值结果，那么你可以做的是：

从每个子数组的最小值开始，
对剩余值进行排序，跟踪其源子阵列
并向上移动已排序的值，直到您达到或超过指定的min

此类功能的一个示例（使用LINQ简洁，并转换回ArrayList）将是这样的：

public static ArrayList GetNums(int min, int max)
{
    var arr = GenerateArraylistValues();

    // Initialize our results with the minimum values from each array.
    var results = arr.Cast<ArrayList>().Select(sub => sub.Cast<int>().Min()).ToList();
    var others = new List<UnionItem>();

    for (int i = 0; i < arr.Count; i++)
    {
        others = new List<UnionItem>(others
                    // Concatenate the arrays together.
                    .Union(((ArrayList)arr[i]).Cast<int>()
                    // Don't need the min value, which we will start at already. (Optional)
                    .Where(val => val != results[i])
                    // Create the UnionItem, to hold the value and the original array source.
                    .Select(val => new UnionItem(val, i))));
    }

    // Order our combined values.
    others = new List<UnionItem>(others.OrderBy(val => val.Value));
    using (var next = others.GetEnumerator())
    {
        // Progress through the combined values until we (a) meet or exceed min, or (b) run out of values.
        while ((results.Sum() < min) && (next.MoveNext()))
        {
            // Update the list of result values according to the UnionItem source.
            results[next.Current.Source] = next.Current.Value;
        }
    }

    // Once through our calculation, check now if we've successfully met the conditions.
    int sum = results.Sum();
    if (sum >= min && sum <= max)
    {
        return new ArrayList(results);
    }
    else
    {
        // Whatever happens if no valid match.
        return new ArrayList();
    }
}

private class UnionItem
{
    public readonly int Value;   // Holds the value from the array.
    public readonly int Source;  // Holds the index of the source array.
    public UnionItem(int value, int source)
    {
        Value = value;
        Source = source;
    }
}

我已经创建了UnionItem类，只是为了我们可以跟踪每个值的源数组，以正确地替换我们的结果。

从2D Arraylist中提取整数，其总和在一个范围之间

2 个答案: