# 238. Product of Array Except self

Source: Internet
Author: User

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]` .

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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The array, in addition to its own elements, is a product of other elements that form a new array.

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Ideas:

Set two integer variables, start=end=1, new return array vector<int> re (Nums.size (), 1);

Multiplies & assigns the current element in the re from the end of the array, respectively

As you traverse one side, each element in the re is exactly what is required.

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Code

`classSolution { Public: Vector<int> productexceptself (vector<int>&nums) {        if(Nums.empty ())returnvector<int>(); Vector<int> Re (nums.size (),1); intStart =1; intE =1;  for(inti =0;i< (int) nums.size (); i++) {Re[i]*=start; Start*=Nums[i]; Re[nums.size ()-i-1] *=e; E*= nums[nums.size ()-i-1]; }         for(Auto I:re) cout<<i<<" "; cout<<Endl; returnre; }};`

238. Product of Array Except self

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