我不喜欢上面看到的Javascript解决方案。下面是我使用部分应用、闭包和递归构建的一个:

好的，我主要关心的是，如果组合数组能满足目标要求，希望这样你就能找到剩下的组合了

这里只需要设置目标并传递组合数组。

function main() {
    const target = 10
    const getPermutationThatSumT = setTarget(target)
    const permutation = getPermutationThatSumT([1, 4, 2, 5, 6, 7])

    console.log( permutation );
}

我提出的当前实现

function setTarget(target) {
    let partial = [];

    return function permute(input) {
        let i, removed;
        for (i = 0; i < input.length; i++) {
            removed = input.splice(i, 1)[0];
            partial.push(removed);

            const sum = partial.reduce((a, b) => a + b)
            if (sum === target) return partial.slice()
            if (sum < target) permute(input)

            input.splice(i, 0, removed);
            partial.pop();
        }
        return null
    };
}

建议回答:

下面是一个使用es2015生成器的解决方案:

function* subsetSum(numbers, target, partial = [], partialSum = 0) {

  if(partialSum === target) yield partial

  if(partialSum >= target) return

  for(let i = 0; i < numbers.length; i++){
    const remaining = numbers.slice(i + 1)
        , n = numbers[i]

    yield* subsetSum(remaining, target, [...partial, n], partialSum + n)
  }

}

使用生成器实际上非常有用，因为它允许您在找到有效子集时立即暂停脚本执行。这与没有生成器(即缺乏状态)的解决方案形成对比，后者必须遍历每个数字子集

我不喜欢上面看到的Javascript解决方案。下面是我使用部分应用、闭包和递归构建的一个:

好的，我主要关心的是，如果组合数组能满足目标要求，希望这样你就能找到剩下的组合了

这里只需要设置目标并传递组合数组。

function main() {
    const target = 10
    const getPermutationThatSumT = setTarget(target)
    const permutation = getPermutationThatSumT([1, 4, 2, 5, 6, 7])

    console.log( permutation );
}

我提出的当前实现

function setTarget(target) {
    let partial = [];

    return function permute(input) {
        let i, removed;
        for (i = 0; i < input.length; i++) {
            removed = input.splice(i, 1)[0];
            partial.push(removed);

            const sum = partial.reduce((a, b) => a + b)
            if (sum === target) return partial.slice()
            if (sum < target) permute(input)

            input.splice(i, 0, removed);
            partial.pop();
        }
        return null
    };
}

c++版本的相同算法

#include <iostream>
#include <list>
void subset_sum_recursive(std::list<int> numbers, int target, std::list<int> partial)
{
        int s = 0;
        for (std::list<int>::const_iterator cit = partial.begin(); cit != partial.end(); cit++)
        {
            s += *cit;
        }
        if(s == target)
        {
                std::cout << "sum([";

                for (std::list<int>::const_iterator cit = partial.begin(); cit != partial.end(); cit++)
                {
                    std::cout << *cit << ",";
                }
                std::cout << "])=" << target << std::endl;
        }
        if(s >= target)
            return;
        int n;
        for (std::list<int>::const_iterator ai = numbers.begin(); ai != numbers.end(); ai++)
        {
            n = *ai;
            std::list<int> remaining;
            for(std::list<int>::const_iterator aj = ai; aj != numbers.end(); aj++)
            {
                if(aj == ai)continue;
                remaining.push_back(*aj);
            }
            std::list<int> partial_rec=partial;
            partial_rec.push_back(n);
            subset_sum_recursive(remaining,target,partial_rec);

        }
}

void subset_sum(std::list<int> numbers,int target)
{
    subset_sum_recursive(numbers,target,std::list<int>());
}
int main()
{
    std::list<int> a;
    a.push_back (3); a.push_back (9); a.push_back (8);
    a.push_back (4);
    a.push_back (5);
    a.push_back (7);
    a.push_back (10);
    int n = 15;
    //std::cin >> n;
    subset_sum(a, n);
    return 0;
}

Javascript版本:

function subsetSum(numbers, target, partial) { var s, n, remaining; partial = partial || []; // sum partial s = partial.reduce(function (a, b) { return a + b; }, 0); // check if the partial sum is equals to target if (s === target) { console.log("%s=%s", partial.join("+"), target) } if (s >= target) { return; // if we reach the number why bother to continue } for (var i = 0; i < numbers.length; i++) { n = numbers[i]; remaining = numbers.slice(i + 1); subsetSum(remaining, target, partial.concat([n])); } } subsetSum([3,9,8,4,5,7,10],15); // output: // 3+8+4=15 // 3+5+7=15 // 8+7=15 // 5+10=15

下面是一个更好的版本，具有更好的输出格式和c++ 11特性:

void subset_sum_rec(std::vector<int> & nums, const int & target, std::vector<int> & partialNums) 
{
    int currentSum = std::accumulate(partialNums.begin(), partialNums.end(), 0);
    if (currentSum > target)
        return;
    if (currentSum == target) 
    {
        std::cout << "sum([";
        for (auto it = partialNums.begin(); it != std::prev(partialNums.end()); ++it)
            cout << *it << ",";
        cout << *std::prev(partialNums.end());
        std::cout << "])=" << target << std::endl;
    }
    for (auto it = nums.begin(); it != nums.end(); ++it) 
    {
        std::vector<int> remaining;
        for (auto it2 = std::next(it); it2 != nums.end(); ++it2)
            remaining.push_back(*it2);

        std::vector<int> partial = partialNums;
        partial.push_back(*it);
        subset_sum_rec(remaining, target, partial);
    }
}