{-
Recursive solution using sqrt(n) subsets. Runs in O(n).

Recursively computes the solution on sqrt(n) subsets of size sqrt(n). 
Then recurses on the product sum of each subset.
Then for each element in each subset, it computes the product with
the product sum of all other products.
Then flattens all subsets.

Recurrence on the run time is T(n) = sqrt(n)*T(sqrt(n)) + T(sqrt(n)) + n

Suppose that T(n) ≤ cn in O(n).

T(n) = sqrt(n)*T(sqrt(n)) + T(sqrt(n)) + n
    ≤ sqrt(n)*c*sqrt(n) + c*sqrt(n) + n
    ≤ c*n + c*sqrt(n) + n
    ≤ (2c+1)*n
    ∈ O(n)

Note that ceiling(sqrt(n)) can be computed using a binary search 
and O(logn) iterations, if the sqrt instruction is not permitted.
-}

otherProducts [] = []
otherProducts [x] = [1]
otherProducts [x,y] = [y,x]
otherProducts a = foldl' (++) [] $ zipWith (\s p -> map (*p) s) solvedSubsets subsetOtherProducts
    where 
      n = length a

      -- Subset size. Require that 1 < s < n.
      s = ceiling $ sqrt $ fromIntegral n

      solvedSubsets = map otherProducts subsets
      subsetOtherProducts = otherProducts $ map product subsets

      subsets = reverse $ loop a []
          where loop [] acc = acc
                loop a acc = loop (drop s a) ((take s a):acc)

这是O(n²)但f#太漂亮了

List.fold (fun seed i -> List.mapi (fun j x -> if i=j+1 then x else x*i) seed) 
          [1;1;1;1;1]
          [1..5]

下面是Ruby中的一行程序解决方案。

全国矿工工会。映射{|n| (num - [n]).inject(:*)}

JavaScript中使用reduce的变体

const getProduct = arr => arr。Reduce ((acc, value) => acc * value); const arrayWithExclusion = (arr, node) => 加勒比海盗。Reduce ((acc, val, j) => (node !== j ?)[…Acc, val]: Acc)， []); const getproductwitheexclusion = arr => { Let result = []; 对于(设I = 0;I < arrr .length;I += 1) { 结果。推动(getProduct (arrayWithExclusion(加勒比海盗,我))); ｝ 返回结果; }；

在这里添加我的javascript解决方案，因为我没有发现任何人建议这样做。 除法是什么，除了数从另一个数中得到一个数的次数吗?我计算了整个数组的乘积，然后遍历每个元素，并减去当前元素直到0:

//No division operation allowed
// keep substracting divisor from dividend, until dividend is zero or less than divisor
function calculateProducsExceptCurrent_NoDivision(input){
  var res = [];
  var totalProduct = 1;
  //calculate the total product
  for(var i = 0; i < input.length; i++){
    totalProduct = totalProduct * input[i];
  }
  //populate the result array by "dividing" each value
  for(var i = 0; i < input.length; i++){
    var timesSubstracted = 0;
    var divisor = input[i];
    var dividend = totalProduct;
    while(divisor <= dividend){
      dividend = dividend - divisor;
      timesSubstracted++;
    }
    res.push(timesSubstracted);
  }
  return res;
}

这是ptyhon版本

  # This solution use O(n) time and O(n) space
  def productExceptSelf(self, nums):
    """
    :type nums: List[int]
    :rtype: List[int]
    """
    N = len(nums)
    if N == 0: return

    # Initialzie list of 1, size N
    l_prods, r_prods = [1]*N, [1]*N

    for i in range(1, N):
      l_prods[i] = l_prods[i-1] * nums[i-1]

    for i in reversed(range(N-1)):
      r_prods[i] = r_prods[i+1] * nums[i+1]

    result = [x*y for x,y in zip(l_prods,r_prods)]
    return result

  # This solution use O(n) time and O(1) space
  def productExceptSelfSpaceOptimized(self, nums):
    """
    :type nums: List[int]
    :rtype: List[int]
    """
    N = len(nums)
    if N == 0: return

    # Initialzie list of 1, size N
    result = [1]*N

    for i in range(1, N):
      result[i] = result[i-1] * nums[i-1]

    r_prod = 1
    for i in reversed(range(N)):
      result[i] *= r_prod
      r_prod *= nums[i]

    return result