Leetcode: Product of Array Except Self

最新推荐文章于 2020-01-07 12:40:53 发布

原创最新推荐文章于 2020-01-07 12:40:53 发布 · 337 阅读

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本文探讨了如何在不使用除法的情况下计算出一个数组中每个元素对应的除自身以外所有元素的乘积，并提供了一个时间复杂度为O(n)的解决方案。此外，还讨论了如何在常数空间复杂度下解决此问题。

Question

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.)

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Solution

Get idea from here, here2

class Solution(object):
    def productExceptSelf(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """

        res = [0]*len(nums)
        n = 1
        for ind,elem in enumerate(nums):
            res[ind] = n
            n *= elem

        n = 1
        for ind in range(len(nums)-1,-1,-1):
            res[ind] *= n
            n *= nums[ind]

        return res