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

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]

将Michael Anderson的解决方案翻译成Haskell:

otherProducts xs = zipWith (*) below above

     where below = scanl (*) 1 $ init xs

           above = tail $ scanr (*) 1 xs

我们正在分解数组的元素，首先从下标之前开始，即前缀，然后是下标或后缀之后

class Solution:

   def productExceptSelf(nums):

      length = len(nums)


      result = [1] * length


      prefix_product = 1


      postfix_product = 1

# we initialize the result and products


      for i in range(length)

      result[i] *= prefix_product


       prefix_product *= nums[i]

#we multiply the result by each number before the index

      for i in range(length-1,-1,-1)

      result[i] *= postfix_product


      postfix_product *= nums[i]

#same for after index
   return result

抱歉，走路时用手机

{-
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)

试试这个!

import java.util.*;
class arrProduct
{
 public static void main(String args[])
     {
         //getting the size of the array
         Scanner s = new Scanner(System.in);
            int noe = s.nextInt();

        int out[]=new int[noe];
         int arr[] = new int[noe];

         // getting the input array
         for(int k=0;k<noe;k++)
         {
             arr[k]=s.nextInt();
         }

         int val1 = 1,val2=1;
         for(int i=0;i<noe;i++)
         {
             int res=1;

                 for(int j=1;j<noe;j++)
                 {
                if((i+j)>(noe-1))
                {

                    int diff = (i+j)-(noe);

                    if(arr[diff]!=0)
                    {
                    res = res * arr[diff];
                    }
                }

                else
                {
                    if(arr[i+j]!=0)
                    {
                    res= res*arr[i+j];
                    }
                }


             out[i]=res;

         }
         }

         //printing result
         System.out.print("Array of Product: [");
         for(int l=0;l<out.length;l++)
         {
             if(l!=out.length-1)
             {
            System.out.print(out[l]+",");
             }
             else
             {
                 System.out.print(out[l]);
             }
         }
         System.out.print("]");
     }

}