Leetcode: Product of Array Except Self (60ms) analysis and solution

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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Analysis:

Using two round processing: forward and backward. See figure for detail:





//////////////////////////////////////////

//code 60ms

class Solution {

public:

vector<int> productExceptSelf(vector<int>& nums) {

vector<int> tmp(nums.size(), 1);

int n=nums.size();

//forward

for(int i=0;i<n-1;i++){

tmp[i+1]=tmp[i]*nums[i];

}

//backward

int m=1;

for (int j=n-1;j>-1;j--){

tmp[j]=m*tmp[j];

m=m*nums[j];

}

return tmp;

}

};