Question

给定一个数组说nums = {1,2,5,3,6，-1，-2,10,11,12}，使用max no of elements（比如maxNums = 3）找到其总和（比如sum = 10）= K

的元素

所以如果要使用的maxNums = 3 总和找到= 10 答案是

    {1  3  6}    
    {1  -1  10}
    {1  -2  11}
    {2  5  3}
    {2  -2  10}
    {5 6 -1}
    {-1  11}
    {-2  12}
    {10}

我写了一个递归函数来完成这项工作。 如何在没有递归的情况下执行此操作？ 和/或记忆力较少？

class Program
{
        static Int32[] nums = { 1,2,5,3,6,-1,-2,10,11,12};
        static Int32 sum = 10;
        static Int32 maxNums = 3;


        static void Main(string[] args)
        {

            Int32[] arr = new Int32[nums.Length];
            CurrentSum(0, 0, 0, arr);

            Console.ReadLine();
        }

        public static void Print(Int32[] arr)
        {
            for (Int32 i = 0; i < arr.Length; i++)
            {
                if (arr[i] != 0)
                    Console.Write("  "   +arr[i]);
            }
            Console.WriteLine();
        }


        public static void CurrentSum(Int32 sumSoFar, Int32 numsUsed, Int32 startIndex, Int32[] selectedNums)
        {
            if ( startIndex >= nums.Length  || numsUsed > maxNums)
            {
                if (sumSoFar == sum && numsUsed <= maxNums)
                {
                    Print(selectedNums);
                }                    
                return;
            }

                       **//Include the next number and check the sum**
                    selectedNums[startIndex] = nums[startIndex];
                    CurrentSum(sumSoFar + nums[startIndex], numsUsed+1, startIndex+1, selectedNums);

                    **//Dont include the next number**
                    selectedNums[startIndex] = 0;
                    CurrentSum(sumSoFar , numsUsed , startIndex + 1, selectedNums);
        }
    }

Answer 1

你的功能看起来很好但可能有点优化：

class Program
{
    static Int32[] nums = { 1, 2, 5, 3, 6, -1, -2, 10, 11, 12 };
    static Int32 sum = 10;
    static Int32 maxNums = 3;
    static Int32[] selectedNums = new Int32[maxNums];

    static void Main(string[] args)
    {
        CurrentSum(0, 0, 0);
        Console.ReadLine();
    }

    public static void Print(int count)
    {
        for (Int32 i = 0; i < count; i++)
        {
            Console.Write(" " + selectedNums[i]);
        }
        Console.WriteLine();
    }

    public static void CurrentSum(Int32 sumSoFar, Int32 numsUsed, Int32 startIndex)
    {
        if (sumSoFar == sum && numsUsed <= maxNums)
        {
            Print(numsUsed);
        }

        if (numsUsed >= maxNums || startIndex >= nums.Length)
            return;

        for (int i = startIndex; i < nums.Length; i++)
        {
            // Include i'th number
            selectedNums[numsUsed] = nums[i];
            CurrentSum(sumSoFar + nums[i], numsUsed + 1, i + 1);
        }
    }
}

我还修复了你的功能中的一个错误。它无法使用以下测试用例：

{10, 2, -2}
Sum = 10
K = 3

您的函数仅返回{10}而不是{10} and {10, 2, -2}

Answer 2

和Haskell解决方案......

import Data.List
listElements max_num k arr = 
  filter (\x -> sum x == k && length x == max_num) $ subsequences arr

*主＆GT; listElements 3 10 [1,2,5,3,6，-1，-2,10,11,12]
[[2,5,3]，[1,3,6]，[5,6，-1]，[1，-1,10]，[2，-2,10-]，[1，-2， 11]

数组中K个元素的总和等于N.

2 个答案: