# Leetcode:product of Array Except self

Source: Internet
Author: User

Problem:

Given an array of n integers where n > 1, `nums` and return an array `output` such that's equal to `output[i]` th E 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 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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`1 classSolution {2  Public:3vector<int> productexceptself (vector<int>&nums) {4         5         /*6 idea: Left[i] represents the product of the number before I7 Right[i] Represents the product of the number after I8 then total[i] = Left[i]*right[i]9 sweep the surface two times array resultsTen         */ One          Avector<int> Left (nums.size (),1); -         -         for(intI=1; I<nums.size (); i++) the        { -left[i]=left[i-1]*nums[i-1]; -        } -         +        intright=1; -         +         for(intI=nums.size ()-2; i>=0; i--) A        { atright=nums[i+1]*Right ; -left[i]*=Right ; -        } -          -        returnLeft ; -          in          -     } to};`

Leetcode:product of Array Except self

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