Product of Array Exclude Itself

题目：

Given an integers array A.

Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B WITHOUT divide operation.

分析：

这题目不允许用除法，就不谈了，如果可以用除法，必须要考虑有没有0的问题，其实更容易中陷阱

同时这个题目有corner case分别是nums长度为0或者为1，长度为1的情况比较容易忘记

解法：

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
        if (nums.size() == 0) {
            return vector<long long>();
        }

        vector<long long> forward(nums.size(), 1);
        vector<long long> backward(nums.size(), 1);

        forward[0] = nums[0];
        for (int i = 1; i < nums.size(); i++) {
            forward[i] = forward[i - 1] * nums[i];
        }

        backward[nums.size() - 1] = nums[nums.size() - 1];
        for (int j = nums.size() - 2; j >= 0; j--) {
            backward[j] = backward[j + 1] * nums[j];
        }

        vector<long long> res(nums.size(), 1);
        if (nums.size() == 1) {
            return res;
        }
        res[0] = backward[1];
        res[nums.size() - 1] = forward[nums.size() - 2];
        for (int i = 1; i < nums.size() - 1; i++) {
            res[i] = forward[i - 1] * backward[i + 1];
        }
        return res;
    }
};