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。

下面是我尝试用Java来解决这个问题。抱歉格式不规范，但代码有很多重复，这是我能做的最好的，使它可读。

import java.util.Arrays;

public class Products {
    static int[] products(int... nums) {
        final int N = nums.length;
        int[] prods = new int[N];
        Arrays.fill(prods, 1);
        for (int
           i = 0, pi = 1    ,  j = N-1, pj = 1  ;
           (i < N)         && (j >= 0)          ;
           pi *= nums[i++]  ,  pj *= nums[j--]  )
        {
           prods[i] *= pi   ;  prods[j] *= pj   ;
        }
        return prods;
    }
    public static void main(String[] args) {
        System.out.println(
            Arrays.toString(products(1, 2, 3, 4, 5))
        ); // prints "[120, 60, 40, 30, 24]"
    }
}

循环不变量为pi = nums[0] * nums[1] *..* nums[N-2] *..num [j + 1]。左边的i部分是“前缀”逻辑，右边的j部分是“后缀”逻辑。

递归一行程序

Jasmeet给出了一个(漂亮的!)递归解;我把它变成了这样(可怕!)Java一行程序。它进行就地修改，堆栈中有O(N)个临时空间。

static int multiply(int[] nums, int p, int n) {
    return (n == nums.length) ? 1
      : nums[n] * (p = multiply(nums, nums[n] * (nums[n] = p), n + 1))
          + 0*(nums[n] *= p);
}

int[] arr = {1,2,3,4,5};
multiply(arr, 1, 0);
System.out.println(Arrays.toString(arr));
// prints "[120, 60, 40, 30, 24]"

O(n)时间的简洁解:

对于每个元素，计算在它之前出现的所有元素的乘积，并将其存储在数组“pre”中。 对于每个元素，计算该元素之后所有元素的乘积，并将其存储在数组“post”中 为元素i创建一个最终数组result， 结果[i] = pre[i-1]*post[i+1];

下面是我用现代c++编写的解决方案。它使用std::transform，很容易记住。

在线代码(wandbox)。

#include<algorithm>
#include<iostream>
#include<vector>

using namespace std;

vector<int>& multiply_up(vector<int>& v){
    v.insert(v.begin(),1);
    transform(v.begin()+1, v.end()
             ,v.begin()
             ,v.begin()+1
             ,[](auto const& a, auto const& b) { return b*a; }
             );
    v.pop_back();
    return v;
}

int main() {
    vector<int> v = {1,2,3,4,5};
    auto vr = v;

    reverse(vr.begin(),vr.end());
    multiply_up(v);
    multiply_up(vr);
    reverse(vr.begin(),vr.end());

    transform(v.begin(),v.end()
             ,vr.begin()
             ,v.begin()
             ,[](auto const& a, auto const& b) { return b*a; }
             );

    for(auto& i: v) cout << i << " "; 
}

这是我的代码:

int multiply(int a[],int n,int nextproduct,int i)
{
    int prevproduct=1;
    if(i>=n)
        return prevproduct;
    prevproduct=multiply(a,n,nextproduct*a[i],i+1);
    printf(" i=%d > %d\n",i,prevproduct*nextproduct);
    return prevproduct*a[i];
}

int main()
{
    int a[]={2,4,1,3,5};
    multiply(a,5,1,0);
    return 0;
}

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