[leetcode]Product of Array Except Self

Product of Array Except Self My Submissions Question
Total Accepted: 21825 Total Submissions: 57345 Difficulty: Medium
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.)

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题意就是求一个数组里面 除了他本身的其他位置的乘积。并且要求不能用到除法，并且时间复杂度O(n),空间复杂度O(1)~

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        vector<int> res(n,1);
        for(int i = 0;i != n - 1;++i){
            res[i + 1] = nums[i] * res[i];//计算前半部分的乘积
        }
        for(int i = n - 1;i > 0;--i){//计算后半部分的乘积
            res[0] *= nums[i];
            if(i - 1 == 0)  return res;
            res[i - 1] *= res[0];
        }
        return res;
    }
};