238. Product of Array Except Self -M

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.)

A:

从前向后走一遍，计算之前的乘积
再从后向前走一遍，计算之后的乘积，

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        vector<int> res(n,1);
        // two round, forward, and backward
        for(int i =1;i<n;++i)
            res[i] = res[i-1] * nums[i-1];
        int back = 1;
        for(int i = n-1; i>=0; --i)
        {
            res[i] *= back;
            back *= nums[i];
        }
        return res;
    }
};

用2个帮助数组

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int pre[] = new int[n];
        int after[] = new int[n];
        pre[0] = 1;
        for(int i =1;i<n;i++)
            pre[i] = pre[i-1]*nums[i-1];
        after[n-1] = 1;
        for(int i=n-2;i>=0;i--)
            after[i] = after[i+1] * nums[i+1];
        int[] result = new int[n];
        for(int i =0;i<n;i++)
            result[i] = pre[i] * after[i];
        return result;
    }
}

Follow up:
为节省空间，用result数组和给的nums数组代替pre   after 数组。

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] pre = new int[n];
        pre[0] = 1;
        for(int i =1;i<n;i++)
            pre[i] = pre[i-1]*nums[i-1];
        for(int i=n-2;i>0;i--)
            nums[i] = nums[i+1] * nums[i];
        for(int i =0;i<n-1;i++)
            pre[i] = pre[i] * nums[i+1];
        return pre;
    }
}

Mistakes: