Google : 计算a[0]*a[1]*…*a[n-1]/a[i]（转）

有空不防看一下这个博客：http://jonnyhsy.wordpress.com/category/algorithms-data-structure/

//Given an array a[n], build another array b[n], b[i] = a[0]*a[1]*…*a[n-1]/a[i]
//no division can be used, O(n) time complexity

见网页：

http://www.ihas1337code.com/2010/04/multiplication-of-numbers.html

Let’s define array B where element B[i] = multiplication of numbers from A[0] to A[i]. For example, if A = {4, 3, 2, 1, 2}, then B = {4, 12, 24, 24, 48}. Then, we scan the array A from right to left, and have a temporary variable called product which stores the multiplication from right to left so far. Calculating OUTPUT[i] is straight forward, as OUTPUT[i] = B[i-1] * product.

The above method requires only O(n) time but uses O(n) space. We have to trade memory for speed. Is there a better way? (i.e., runs in O(n) time but without extra space?)

Yes, actually the temporary table is not required. We can have two variables called left and right, each keeping track of the product of numbers multiplied from left->right and right->left. Could you see why this works without extra space?

void array_multiplication(int A[], int OUTPUT[], int n)
{
 int left = 1;
 int right = 1;
  for (int i = 0; i < n; i++)
  OUTPUT[i] = 1;
  for (int i = 0; i < n; i++) {
    OUTPUT[i] *= left;
    OUTPUT[n - 1 - i] *= right;
    left *= A[i];
    right *= A[n - 1 - i];
  }
}

载自：http://jonnyhsy.wordpress.com/2011/05/03/google-%E8%AE%A1%E7%AE%97a0a1-an-1ai/