[leetcode] 238. Product of Array Except Self @ python

版权声明：版权归个人所有，未经博主允许，禁止转载 https://blog.csdn.net/danspace1/article/details/86601341

原题

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.)

解法

由于不能用除法, 我们用两趟遍历, 第一趟从左到右, 计算在index i左边的数的乘积, 第二趟从右到左, 将index右边的乘积乘到左边. 例如nums = [1, 2, 3, 4],
第一趟遍历后的结果为[1, 1, 2, 6], 第二趟遍历时, p初始值为1, 然后将nums右边的数累积乘到p, 遍历结束后结果为[24, 12, 8, 6]
Time: 2*O(n)
Space: O(1)

代码

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