根据结果和集合长度获取原数组中所有符合的集合

// 计算数字之和

const nums = [-1, 0, 1, 2, 3]

function countNums(num, nums, length) {
    function getSum(data) {
        let sum = 0;
        for (let value of data) {
            sum += value;
        }
        return sum
    }
    function getNumbers(source, count, isPermutation = true) {
        //如果只取一位，返回数组中的所有项，例如 [ [1], [2], [3] ]
        let currentList = source.map((item) => [item]);
        if (count === 1) {
            return currentList;
        }
        let result = [];
        //取出第一项后，再取出后面count - 1 项的排列组合，并把第一项的所有可能（currentList）和 后面count-1项所有可能交叉组合
        for (let i = 0; i < currentList.length; i++) {
            let current = currentList[i];
            //如果是排列的方式，在取count-1时，源数组中排除当前项
            let children = [];
            if (isPermutation) {
                children = getNumbers(source.filter(item => item !== current[0]), count - 1, isPermutation);
            } else {
                //如果是组合的方法，在取count-1时，源数组只使用当前项之后的
                children = getNumbers(source.slice(i + 1), count - 1, isPermutation);
            }
            for (let child of children) {
                result.push([...current, ...child]);
            }
        }
        return result;
    }
    const result = getNumbers(nums, length, false)
    const nRes = []
    for (const numbers of result) {
        const sum = getSum(numbers)
        if (sum === num) {
            nRes.push(numbers)
        }
    }
    return nRes
}

const result = countNums(2, nums, 3)
console.log(result)