leetcode - Product of Array Except Self

leetcode - Product of Array Except Self

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 of nums 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) {
        vector<int> totalProductForward(nums.size());
        totalProductForward[0] = nums[0];
        for(int i = 1; i < nums.size(); i++){
            totalProductForward[i]= totalProductForward[i-1]*nums[i];
        }
        vector<int> totalProductBackward(nums.size());
        totalProductBackward[nums.size()-1] = nums[nums.size()-1];
        for(int j = nums.size()-2; j >= 0; j--){
            totalProductBackward[j] = totalProductBackward[j+1]*nums[j];
        }
        vector<int> res(nums.size());
        res[0] = totalProductBackward[1];
        res[nums.size()-1] = totalProductForward[nums.size()-2];
        for(int j = 1; j < nums.size()-1; j++){
            res[j] = (totalProductBackward[j+1] * totalProductForward[j-1]);
        }
        return res;
    }

};

空间复杂度O(3n).分别向后和向前遍历一遍，求出累乘的积的数组，然后每个结果都可以由前一部分积和后一部分积相乘获得。