LeetCode:Product of Array Except Self

Problem:

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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class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        
        /*
        思路:left[i]表示i之前的数的乘积  
        right[i]表示i之后的数的乘积 
        那么total[i] = left[i]*right[i]
        扫面两遍数组出结果
        */
        
       vector<int> left(nums.size(),1);
       
       for(int i=1;i<nums.size();i++)
       {
           left[i]=left[i-1]*nums[i-1];
       }
       
       int right=1;
       
       for(int i=nums.size()-2;i>=0;i--)
       {
           right=nums[i+1]*right;
           left[i]*=right;
       }
        
       return left;
        
        
    }
};