#50 Product of Array Exclude Itself

題目描述：

Given an integers array A.

Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate BWITHOUT divide operation.

Have you met this question in a real interview? 

Yes

Example

For A = [1, 2, 3], return [6, 3, 2].

題目思路：

這題用一個left數組去記錄i往左所有數的乘積；right數組去記錄i往右所有數的乘積。然後重新遍歷一遍數組，每個i的答案就是left[i - 1] * right[i + 1]。

Mycode（AC = 12ms）：

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) {
        // write your code here
        vector<long long> ans;
        if (nums.size() == 0) return ans;
        else if (nums.size() == 1) {
            ans.push_back(1);
            return ans;
        }
        
        ans.resize(nums.size(), 1);
        vector<long long> left(nums.size(), nums[0]), right(nums.size(), nums[nums.size() - 1]);
        
        // left[i] denote the product of all numbers
        // from 0...i
        for (int i = 1; i < nums.size(); i++) {
            left[i] = left[i - 1] * nums[i];
        }
        
        // right[i] denote the product of all numbers
        // from i...nums.size() - 1
        for (int i = nums.size() - 2; i >= 0; i--) {
            right[i] = right[i + 1] * nums[i];
        }
        
        for (int i = 0; i < nums.size(); i++) {
            if (i == 0) {
                ans[i] = right[i + 1];
            }
            else if (i == nums.size() - 1) {
                ans[i] = left[i - 1];
            }
            else {
                ans[i] = right[i + 1] * left[i - 1];
            }
        }
        
        return ans;
    }
};