Leetcode 238 Product of Array Except Self

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).

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

Follow up:
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.)

两次循环,第一次循环得出前面所有元素之积,第二次乘以后面所有元素之积。

class Solution:

    # @param {integer[]} nums

    # @return {integer[]}

    def productExceptSelf(self, nums):

        result = [0 for x in range(len(nums))]

        before = after = 1

        for i in range(len(nums)):

            result[i] = before

            before *= nums[i]

        for i in range(len(nums)-1,-1,-1):

            result[i] *= after

            after *= nums[i]

        return result

 

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