[Facebook] Products of all elements

快速计算数组中除自身外其他元素乘积的方法

最新推荐文章于 2025-10-08 14:05:18 发布

原创最新推荐文章于 2025-10-08 14:05:18 发布 · 220 阅读

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本文介绍了一种O(N)复杂度的算法，用于计算数组中每个元素除了自身以外所有元素的乘积，无需使用除法运算。

Given an array of numbers, nums, return an array of numbers products, where products[i] is the product of all nums[j], j != i.

Input : [1, 2, 3, 4, 5]

Output: [(2*3*4*5), (1*3*4*5), (1*2*4*5), (1*2*3*5), (1*2*3*4)]

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

You must do this in O(N) without using division.

[Thoughts]

An explaination of polygenelubricants method is: The trick is to construct the arrays (in the case for 4 elements)

{              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,  }

Both of which can be done in O(n) by starting at the left and right edges respectively.
Then multiplying the two arrays element by element gives the required result
My code would look something like this:

int a[N] // This is the input
int products_below[N];
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];
}

If you need to be O(1) in space too you can do this (which is less clear IMHO)

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

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

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