[Leetcode]238. Product of Array Except Self

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本文介绍了一种算法，可以在O(n)的时间复杂度内计算出数组中每个元素对应的除自身之外所有元素的乘积，且不使用除法运算。通过预计算左侧元素的累积乘积和右侧元素的累积乘积，最终得到结果。

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Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements ofnums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up:

Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        if (nums.empty())
            return {};
        vector<int> leftpro(nums.size());
        leftpro[0] = 1;
        for (int i = 1; i < nums.size(); ++i) {
            leftpro[i] = nums[i - 1] * leftpro[i - 1];
        }
        
        int rightpro = 1;
        for (int i = static_cast<int>(nums.size()) - 2; i >= 0; --i) {
            rightpro *= nums[i + 1];
            leftpro[i] = leftpro[i] * rightpro;
        }
        return leftpro;    
    }
};