238. Product Of Array Except Self

238. Product Of Array Except Self (Medium) 除自身以外数组的乘积

题干

Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

Example 1:

Input: nums = [1,2,3,4] Output: [24,12,8,6]

Example 2:

Input: nums = [-1,1,0,-3,3] Output: [0,0,9,0,0]

Constraints:

• 2 <= nums.length <= 105
• -30 <= nums[i] <= 30
• The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)

解题思路

朴素算法

answer中的元素，为左半部分（前缀）和右半部分（后缀）。其中左半部分，可以通过循环线性逐个算出，而右半部分，可以倒循环的时候逐个算出。那么先正循环一次，把前缀逐个记住；再倒循环一次，把后缀逐个算出，然后和对应的前缀相乘即可得到结果。

• 时间复杂度：O(n)
• 空间复杂度：O(n)

优化算法

时间复杂度已经无法优化，那么只能优化空间复杂度到O(1)。由于answer是不计算到空间复杂度中的，所以要它身上下手

答案

朴素答案

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        answer = [1] * len(nums)
        prefix_product_list = [1] * len(nums)
        
        for i in range(1, len(nums)):
            prefix_product_list[i] = prefix_product_list[i-1] * nums[i-1]

        suffix_product = 1
        for j in range(len(nums)-1, -1, -1):
            answer[j] = prefix_product_list[j] * suffix_product
            suffix_product *= nums[j]

        return answer

优化答案

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        answer = [1] * len(nums)
        for i in range(1, len(nums)):
            answer[i] = answer[i-1] * nums[i-1]

        suffix_product = 1
        for j in range(len(nums)-1, -1, -1):
            answer[j] *= suffix_product
            suffix_product *= nums[j]

        return answer