ruby的解决方案

a = [1,2,3,4]
result = []
a.each {|x| result.push( (a-[x]).reject(&:zero?).reduce(:*)) }
puts result

def products(nums):
    prefix_products = []
    for num in nums:
        if prefix_products:
            prefix_products.append(prefix_products[-1] * num)
        else:
            prefix_products.append(num)

    suffix_products = []
    for num in reversed(nums):
        if suffix_products:
            suffix_products.append(suffix_products[-1] * num)
        else:
            suffix_products.append(num)
        suffix_products = list(reversed(suffix_products))

    result = []
    for i in range(len(nums)):
        if i == 0:
            result.append(suffix_products[i + 1])
        elif i == len(nums) - 1:
            result.append(prefix_products[i-1])
        else:
            result.append(
                prefix_products[i-1] * suffix_products[i+1]
            )
    return result

最近有人问我这个问题，虽然我不能得到O(N)，但我有一个不同的方法(不幸的是O(N²))，但我想无论如何都要分享。

首先转换为列表<Integer>。

遍历原始数组array.length()次。

使用while循环乘下一组所需的数字:

while (temp < list.size() - 1) {
    res *= list.get(temp);
    temp++;
}

然后将res添加到一个新数组(当然，您已经在前面声明了)，然后将数组[i]的值添加到List，依此类推。

我知道这不会有太大的用处，但这是我在面试的压力下想到的:)

    int[] array = new int[]{1, 2, 3, 4, 5};
    List<Integer> list = Arrays.stream(array).boxed().collect(Collectors.toList());
    int[] newarray = new int[array.length];
    int res = 1;
    for (int i = 0; i < array.length; i++) {
        int temp = i;
        while (temp < list.size() - 1) {
            res *= list.get(temp);
            temp++;
        }
        newarray[i] = res;
        list.add(array[i]);
        res = 1;
    }

输出:[24,120,60,40,30]

我的第一次尝试，用Python。O (2 n):

def product(l):
    product = 1
    num_zeroes = 0
    pos_zero = -1

    # Multiply all and set positions
    for i, x in enumerate(l):
        if x != 0:
            product *= x
            l[i] = 1.0/x
        else:
            num_zeroes += 1
            pos_zero = i

    # Warning! Zeroes ahead!
    if num_zeroes > 0:
        l = [0] * len(l)

        if num_zeroes == 1:
            l[pos_zero] = product

    else:
        # Now set the definitive elements
        for i in range(len(l)):
            l[i] = int(l[i] * product)

    return l


if __name__ == "__main__":
    print("[0, 0, 4] = " + str(product([0, 0, 4])))
    print("[3, 0, 4] = " + str(product([3, 0, 4])))
    print("[1, 2, 3] = " + str(product([1, 2, 3])))
    print("[2, 3, 4, 5, 6] = " + str(product([2, 3, 4, 5, 6])))
    print("[2, 1, 2, 2, 3] = " + str(product([2, 1, 2, 2, 3])))

输出:

[0, 0, 4] = [0, 0, 0]
[3, 0, 4] = [0, 12, 0]
[1, 2, 3] = [6, 3, 2]
[2, 3, 4, 5, 6] = [360, 240, 180, 144, 120]
[2, 1, 2, 2, 3] = [12, 24, 12, 12, 8]

还有一个O(N^(3/2))非最优解。不过，这很有趣。

首先预处理大小为N^0.5的每个部分乘法(这在O(N)时间复杂度中完成)。然后，计算每个数字的其他值的倍数可以在2*O(N^0.5)时间内完成(为什么?因为您只需要将其他((N^0.5) - 1)数字的最后一个元素相乘，并将结果与属于当前数字组的((N^0.5) - 1)数字相乘。对每一个数都这样做，可以得到O(N^(3/2))时间。

例子:

4, 6, 7, 2, 3, 1, 9, 5, 8

部分结果: 4*6*7 = 168 2*3*1 = 6 9*5*8 = 360

要计算3的值，需要将其他组的值乘以168*360，然后乘以2*1。

以下是线性O(n)时间内的简单Scala版本:

def getProductEff(in:Seq[Int]):Seq[Int] = {

   //create a list which has product of every element to the left of this element
   val fromLeft = in.foldLeft((1, Seq.empty[Int]))((ac, i) => (i * ac._1, ac._2 :+ ac._1))._2

   //create a list which has product of every element to the right of this element, which is the same as the previous step but in reverse
   val fromRight = in.reverse.foldLeft((1,Seq.empty[Int]))((ac,i) => (i * ac._1,ac._2 :+ ac._1))._2.reverse

   //merge the two list by product at index
   in.indices.map(i => fromLeft(i) * fromRight(i))

}

这是可行的，因为本质上答案是一个数组，它是左右所有元素的乘积。