LeetCode238 Product of Array Except Self 除自身以外数组的乘积

1. problem description

Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Example:

Input: [1,2,3,4]
Output: [24,12,8,6]

Note: Please solve it without division and in O(n).

Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

2. solution

O(n) time and O(n) space solution:

class Solution {
public:
    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;
    }
};

We just need to change the two vectors to two integers and note that we should do multiplying operations for two related elements of the results vector in each loop:

class Solution {
public:
    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;
    }
};