238. 除了自身以外数组的乘积 - 力扣（LeetCode）

方法 1：暴力法

时间复杂度：O(n²) ❌ 不符合题目要求
空间复杂度：O(1)（不计输出数组）

python 复制代码

# encoding = utf-8
# 开发者：Alen
# 开发时间： 16:25 
# "Stay hungry，stay foolish."


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

        n = len(nums)
        answer = []
        for i in range(n):
            prod = 1
            for j in range(n):
                if j != i:
                    prod *= nums[j]
            answer.append(prod)
        return answer

方法 2：左右乘积列表

时间复杂度：O(n) ✅
空间复杂度：O(n)（用了两个额外数组）

python 复制代码

# encoding = utf-8
# 开发者：Alen
# 开发时间： 16:25 
# "Stay hungry，stay foolish."


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

        n = len(nums)
        left = [1] * n
        right = [1] * n
        for i in range(1, n):
            left[i] = left[i -1] * nums[i - 1]

        for i in range(n - 2, -1, -1):
            right[i] = right[i + 1] * nums[i + 1]

        res = []
        for i in range(n):
            res.append(left[i] * right[i])

        return res

逐步分析：

1. 初始化

复制代码

n = len(nums)
left = [1] * n    # [1, 1, 1, 1]
right = [1] * n   # [1, 1, 1, 1]

2. 计算 left 数组（从左到右）

复制代码

# i=1: left[1] = left[0] * nums[0] = 1 * 1 = 1
# i=2: left[2] = left[1] * nums[1] = 1 * 2 = 2
# i=3: left[3] = left[2] * nums[2] = 2 * 3 = 6
# left = [1, 1, 2, 6]

3. 计算 right 数组（从右到左）

复制代码

# i=2: right[2] = right[3] * nums[3] = 1 * 4 = 4
# i=1: right[1] = right[2] * nums[2] = 4 * 3 = 12
# i=0: right[0] = right[1] * nums[1] = 12 * 2 = 24
# right = [24, 12, 4, 1]

4. 合并结果

复制代码

result = []
result[0] = left[0] * right[0] = 1 * 24 = 24
result[1] = left[1] * right[1] = 1 * 12 = 12
result[2] = left[2] * right[2] = 2 * 4 = 8
result[3] = left[3] * right[3] = 6 * 1 = 6
# result = [24, 12, 8, 6]

方法 3：空间优化版

时间复杂度：O(n) ✅
空间复杂度：O(1) ✅

python 复制代码

# encoding = utf-8
# 开发者：Alen
# 开发时间： 16:25 
# "Stay hungry，stay foolish."


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

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


        R = 1
        for i in range(n - 1, -1, -1):
            answer[i] *= R
            R *= nums[i]

        return answer

结果

解题步骤：https://www.bilibili.com/video/BV1GbrCBmEqN/?vd_source=15b4bc8968fa5203cc470cb68ff72c96