Question

给定一个整数数组和一个整数k，编写一个函数来检查数组中是否存在一组数字，它们的总和等于k。

让我对这个问题感兴趣的是，可以从数组中的两个以上数字中求和。

例如：

Arr [5，4，3，7]
k = 16

输出：true // 5 + 7 + 4 = 16

Answer 1

请注意，此回答仅对此类问题有用。不要忽略查看其他用户的提示

    private int[] givenArrayOfIntegers = new int[]{5, 4, 3, 7};

    private void find(List<Integer> passedIndexes, int fromIndex, int requiredSum) {

        // used java 8th stream opportunities.
        // Can be replaced with good old for loop
        int sumOfThePreviousElements = passedIndexes.stream()
                .mapToInt(i -> givenArrayOfIntegers[i]).sum();

        // iterate over leftover elements
        for (int i = fromIndex; i < givenArrayOfIntegers.length; i++) {
            if (sumOfThePreviousElements + givenArrayOfIntegers[i] == requiredSum) {
                // we've found the subset

                // here I just print values to the output,
                // your preferences may be different
                passedIndexes.forEach(k -> System.out.print(givenArrayOfIntegers[k] + ", "));
                System.out.println(givenArrayOfIntegers[i]);
            } else if (sumOfThePreviousElements + givenArrayOfIntegers[i] < requiredSum) {
                // if sum of the elements less than required try to call this method again
                // with complemented 'i' value
                passedIndexes.add(i);

                // (RECURSION IS HERE)
                find(passedIndexes, i + 1, requiredSum);

                // remove for the next loop purposes
                passedIndexes.remove(passedIndexes.size() - 1);
            }
        }
    }

Answer 2

我从下面的链接中找到了

https://www.geeksforgeeks.org/perfect-sum-problem-print-subsets-given-sum/

您可以尝试一下吗？

class Program
{
    static bool[][] boolean;
    static string Output = "";
    static int Sum1 = 0;

    static void Main(string[] args)
    {

        List<int> abc = new List<int>();
        int[] arr = { 5, 4, 3, 7 };
        int n = arr.Length;
        int sum = 16;
        bool[][] ServicePoint = new bool[n][];
        for (var i = 0; i < ServicePoint.Length; i++)
        {
            ServicePoint[i] = new bool[sum + 1];
        }
        boolean = ServicePoint;
        int z = 0;
        while (z < n)
        {
            boolean[z][0] = true;
            ++z;
        }
        boolean[0][arr[0]] = arr[0] <= sum ? true : false;
        for (int i = 1; i < n; ++i)
        {
            for (int j = 0; j < sum + 1; ++j)
            {
                if (arr[i] <= j)
                {
                    boolean[i][j] = boolean[i - 1][j] || boolean[i - 1][j - arr[i]];
                }
                else
                {
                    boolean[i][j] = boolean[i - 1][j];
                }
            }
        }
        displayOutput(arr, n - 1, sum, abc);

        Console.ReadLine();
    }
    static void displayOutput(int[] arr, int i, int sum, List<int> abc)
    {
        if (i == 0 && sum == 0)
        {
            Output = String.Join(" + ", abc);
            Sum1 = abc.Sum();
            Console.WriteLine("Sum of numbers: " + Output + " = " + Sum1);
            Console.WriteLine("-----------------------------------");
            abc.Clear();
            return;
        }
        if (i == 0 && sum != 0 && boolean[0][sum])
        {
            abc.Add(arr[i]);
            Output = String.Join(" + ", abc);
            Sum1 = abc.Sum();

            Console.WriteLine("Sum of numbers: " + Output + " = " + Sum1);
            Console.WriteLine("-----------------------------------");
            abc.Clear();
            return;
        }
        if (boolean[i - 1][sum])
        {
            List<int> b = new List<int>();
            b.AddRange(abc);
            displayOutput(arr, i - 1, sum, b);

        }
        if (sum >= arr[i] && boolean[i - 1][sum - arr[i]])
        {
            abc.Add(arr[i]);
            displayOutput(arr, i - 1, sum - arr[i], abc);
        }

    }

}

数组中多个数字的总和等于k

2 个答案: