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

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

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; }；

我们可以先从列表中排除nums[j](其中j != i)，然后得到其余部分的乘积;下面是python解决这个难题的方法:

from functools import reduce
def products(nums):
    return [ reduce(lambda x,y: x * y, nums[:i] + nums[i+1:]) for i in range(len(nums)) ]
print(products([1, 2, 3, 4, 5]))

[out]
[120, 60, 40, 30, 24]

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;
}

下面是另一个简单的概念，可以解决O(N)中的问题。

        int[] arr = new int[] {1, 2, 3, 4, 5};
        int[] outArray = new int[arr.length]; 
        for(int i=0;i<arr.length;i++){
            int res=Arrays.stream(arr).reduce(1, (a, b) -> a * b);
            outArray[i] = res/arr[i];
        }
        System.out.println(Arrays.toString(outArray));