Leetcode 238. Product of Array E

文章作者：Tyan
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1. Description

Product of Array Except Self

2. Solution

解析：可以使用两个数组left, right分别保存从左边到第i个元素的积以及从右边到第i个元素的积，则结果中result[i]=left[i-1] * right[i+1]，这种情况下的空间复杂度为O(n)。另一种方式是，计算左边元素积的同时更新结果数组，计算右边积的同时也更新结果数组，这种情况下的空间复杂度为O(1)。

• Version 1

class Solution:
    def productExceptSelf(self, nums):
        length = len(nums)
        result = [1] * length
        left = [1] * length
        right = [1] * length
        left[0] = nums[0]
        right[-1] = nums[-1]

        for i in range(length):
            if i > 0:
                left[i] = left[i - 1] * nums[i]
            if i < length - 1:
                right[-i - 2] = nums[-i - 2] * right[-i - 1]

        for i in range(length):
            if i == 0:
                result[i] = right[i + 1]
            elif i == length - 1:
                result[i] = left[i - 1]
            else:
                result[i] = left[i - 1] * right[i + 1]

        return result

• Version 2

class Solution:
    def productExceptSelf(self, nums):
        length = len(nums)
        result = [1] * length
        left = 1
        right = 1

        for i in range(length - 1):
            left *= nums[i]
            result[i + 1] = result[i + 1] * left

        for i in range(length - 1, 0, -1):
            right *= nums[i]
            result[i - 1] = result[i - 1] * right

        return result

Reference

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