238. Product of Array Except Self (后两种方法有待进一步研究)

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].
方法一、

vector<int> productExceptSelf(vector<int>& nums) {        int len = nums.size();        vector<int> left,right(len),ret(len);        left.push_back(1);        for(int i=1; i<len; i++){            left.push_back(left[i-1]*nums[i-1]);        }        right[len-1]=1;        for(int i=len-2; i>=0; i--){            right[i]=right[i+1]*nums[i+1];        }        for(int i=0; i<len; i++){            ret[i]=left[i]*right[i];        }        return ret;    }

方法二、O(n) time and O(n) space solution.

vector<int> productExceptSelf(vector<int>& nums) {        int n=nums.size();        vector<int> fromBegin(n);        fromBegin[0]=1;        vector<int> fromLast(n);        fromLast[0]=1;        for(int i=1;i<n;i++){            fromBegin[i]=fromBegin[i-1]*nums[i-1];            fromLast[i]=fromLast[i-1]*nums[n-i];        }        vector<int> res(n);        for(int i=0;i<n;i++){            res[i]=fromBegin[i]*fromLast[n-1-i];        }        return res;    }

方法三、

vector<int> productExceptSelf(vector<int>& nums) {        int n=nums.size();        int fromBegin=1;        int fromLast=1;        vector<int> res(n,1);        for(int i=0;i<n;i++){            res[i]*=fromBegin;            fromBegin*=nums[i];            res[n-1-i]*=fromLast;            fromLast*=nums[n-1-i];        }        return res;    }