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]

多基因润滑剂方法的一个解释是:

诀窍是构造数组(在4个元素的情况下):

{              1,         a[0],    a[0]*a[1],    a[0]*a[1]*a[2],  }
{ a[1]*a[2]*a[3],    a[2]*a[3],         a[3],                 1,  }

这两种方法都可以在O(n)中分别从左右边开始。

然后，将两个数组逐个元素相乘，得到所需的结果。

我的代码看起来是这样的:

int a[N] // This is the input
int products_below[N];
int p = 1;
for (int i = 0; i < N; ++i) {
    products_below[i] = p;
    p *= a[i];
}

int products_above[N];
p = 1;
for (int i = N - 1; i >= 0; --i) {
    products_above[i] = p;
    p *= a[i];
}

int products[N]; // This is the result
for (int i = 0; i < N; ++i) {
    products[i] = products_below[i] * products_above[i];
}

如果你也需要空间中的解是O(1)，你可以这样做(在我看来不太清楚):

int a[N] // This is the input
int products[N];

// Get the products below the current index
int p = 1;
for (int i = 0; i < N; ++i) {
    products[i] = p;
    p *= a[i];
}

// Get the products above the current index
p = 1;
for (int i = N - 1; i >= 0; --i) {
    products[i] *= p;
    p *= a[i];
}

    int[] arr1 = { 1, 2, 3, 4, 5 };
    int[] product = new int[arr1.Length];              

    for (int i = 0; i < arr1.Length; i++)
    {
        for (int j = 0; j < product.Length; j++)
        {
            if (i != j)
            {
                product[j] = product[j] == 0 ? arr1[i] : product[j] * arr1[i];
            }
        }
    }

下面是一个C实现 O(n)时间复杂度。 输入

#include<stdio.h>
int main()
{
    int x;
    printf("Enter The Size of Array : ");
    scanf("%d",&x);
    int array[x-1],i ;
    printf("Enter The Value of Array : \n");
      for( i = 0 ; i <= x-1 ; i++)
      {
          printf("Array[%d] = ",i);
          scanf("%d",&array[i]);
      }
    int left[x-1] , right[x-1];
    left[0] = 1 ;
    right[x-1] = 1 ;
      for( i = 1 ; i <= x-1 ; i++)
      {
          left[i] = left[i-1] * array[i-1];
      }
    printf("\nThis is Multiplication of array[i-1] and left[i-1]\n");
      for( i = 0 ; i <= x-1 ; i++)
      {
        printf("Array[%d] = %d , Left[%d] = %d\n",i,array[i],i,left[i]);
      }
      for( i = x-2 ; i >= 0 ; i--)
      {
          right[i] = right[i+1] * array[i+1];
      }
   printf("\nThis is Multiplication of array[i+1] and right[i+1]\n");
      for( i = 0 ; i <= x-1 ; i++)
      {
        printf("Array[%d] = %d , Right[%d] = %d\n",i,array[i],i,right[i]);
      }
    printf("\nThis is Multiplication of Right[i] * Left[i]\n");
      for( i = 0 ; i <= x-1 ; i++)
      {
          printf("Right[%d] * left[%d] = %d * %d = %d\n",i,i,right[i],left[i],right[i]*left[i]);
      }
    return 0 ;
}

输出

    Enter The Size of Array : 5
    Enter The Value of Array :
    Array[0] = 1
    Array[1] = 2
    Array[2] = 3
    Array[3] = 4
    Array[4] = 5

    This is Multiplication of array[i-1] and left[i-1]
    Array[0] = 1 , Left[0] = 1
    Array[1] = 2 , Left[1] = 1
    Array[2] = 3 , Left[2] = 2
    Array[3] = 4 , Left[3] = 6
    Array[4] = 5 , Left[4] = 24

    This is Multiplication of array[i+1] and right[i+1]
    Array[0] = 1 , Right[0] = 120
    Array[1] = 2 , Right[1] = 60
    Array[2] = 3 , Right[2] = 20
    Array[3] = 4 , Right[3] = 5
    Array[4] = 5 , Right[4] = 1

    This is Multiplication of Right[i] * Left[i]
    Right[0] * left[0] = 120 * 1 = 120
    Right[1] * left[1] = 60 * 1 = 60
    Right[2] * left[2] = 20 * 2 = 40
    Right[3] * left[3] = 5 * 6 = 30
    Right[4] * left[4] = 1 * 24 = 24

    Process returned 0 (0x0)   execution time : 6.548 s
    Press any key to continue.

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]

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]