238. Product of Array Except Self（python+cpp）

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_21275321/article/details/84302683

题目：

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.)

解释：’
我们以一个4个元素的数组为例，nums=[a1, a2, a3, a4]。
想在O(n)时间复杂度完成最终的数组输出，res=[a2*a3*a4, a1*a3*a4, a1*a2*a4, a2*a3*a4]。
比较好的解决方法是构造两个数组相乘：
[1, a1, a1*a2, a1*a2*a3]===>list_a
[a2*a3*a4, a3*a4, a4, 1]
python代码：

class Solution(object):
    def productExceptSelf(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        n=len(nums)
        p=1
        list_a=[1]*n
        for i in range(n):
            list_a[i]=p
            p=p*nums[i]
        p=1
        for i in range(n-1,-1,-1):
            list_a[i]*=p
            p=p*nums[i]
        return list_a

c++代码：

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n=nums.size();
        vector<int>a_list(n,1);
        int p=1;
        for (int i=0;i<n;i++)
        {
            a_list[i]=p;
            p*=nums[i];
        }
        p=1;
        for (int i=n-1;i>=0;i--)
        {
            a_list[i]*=p;
            p*=nums[i];
        }
        return a_list;
    }
};

总结：