Given an integers array A.
Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B WITHOUT divide operation.
Example
For A = [1, 2, 3], return [6, 3, 2].
LeetCode上的原题，请参见我之前的博客Product of Array Except Self。
解法一：
class Solution { public: /** * @param A: Given an integers array A * @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1] */ vector<long long> productExcludeItself(vector<int> &nums) { int n = nums.size(); vector<long long> fwd(n, 1), bwd(n, 1), res(n); for (int i = 0; i < n - 1; ++i) { fwd[i + 1] = fwd[i] * nums[i]; } for (int i = n - 1; i > 0; --i) { bwd[i - 1] = bwd[i] * nums[i]; } for (int i = 0; i < n; ++i) { res[i] = fwd[i] * bwd[i]; } return res; } };
解法二：
class Solution { public: /** * @param A: Given an integers array A * @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1] */ vector<long long> productExcludeItself(vector<int> &nums) { long long n = nums.size(), right = 1; vector<long long> res(n, 1); for (int i = 1; i < n; ++i) { res[i] = res[i - 1] * nums[i - 1]; } for (int i = n - 1; i >= 0; --i) { res[i] *= right; right *= nums[i]; } return res; } };
本文转自博客园Grandyang的博客，原文链接：除本身之外的数组之积[LintCode] Product of Array Except Self ，如需转载请自行联系原博主。
