Product of Array Except Self解题报告

Description:

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).

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.)

Example:

For example, given [1,2,3,4], return [24,12,8,6].

Link:

https://leetcode.com/problems/product-of-array-except-self/#/description

解题方法：

扫描两次数组：
第一次从左往右扫描，记录从0 ~ i-1的积。
第二次从右往左扫描，记录从n-1 ~ i+1的积。
将两次扫描的结果对应的位置相乘得到所求的数组。

Tips:

如果出了输出数组不能用额外的空间。则可以用输出数组记录两次扫描的结果。并在第二次扫描的时候将结果算出来。

Time Complexity：

时间O(N) 空间O(1)

完整代码：

vector productExceptSelf(vector& nums) 
    {
        int len = nums.size();
        vector result;
        if(len == 0)
            return result;
        int sum = 1;
        for(int i = 0; i < len; i++)
        {
            result.push_back(sum);
            sum *= nums[i];
        }
        for(int i = len-1, sum = 1; i >= 0; i--)
        {
            result[i] *= sum;  //直接算出结果
            sum *= nums[i];
        }
        return result;
    }