鱼C论坛

 找回密码
 立即注册
查看: 1172|回复: 0

[技术交流] 【朱迪的LeetCode刷题笔记】238. Product of Array Except Self #Medium #Python #C++

[复制链接]
发表于 2023-6-3 12:17:03 | 显示全部楼层 |阅读模式

马上注册,结交更多好友,享用更多功能^_^

您需要 登录 才可以下载或查看,没有账号?立即注册

x
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.

[i][i]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.)

Judy
Python
class Solution(object):
    def productExceptSelf(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        a = 1
        b = 1
        l = len(nums)
        answer = []
        for x in nums:
            answer.append(a)
            a *= x
        for i in range(l-1, -1, -1):
            answer[i] *= b
            b *= nums[i]
        return answer

Mike
C++
vector<int> productExceptSelf(vector<int>& nums) {
        vector<int> result;
        int prod = 1;
        for (auto i : nums) {
            result.push_back(prod);
            prod = i;
        }
        prod = 1;
        for (int i = nums.size() - 1; i >= 0; i--) {
            result[i]= prod;
            prod *= nums[i];
        }
        return result;
    }

本帖被以下淘专辑推荐:

想知道小甲鱼最近在做啥?请访问 -> ilovefishc.com
回复

使用道具 举报

您需要登录后才可以回帖 登录 | 立即注册

本版积分规则

小黑屋|手机版|Archiver|鱼C工作室 ( 粤ICP备18085999号-1 | 粤公网安备 44051102000585号)

GMT+8, 2024-11-18 05:28

Powered by Discuz! X3.4

© 2001-2023 Discuz! Team.

快速回复 返回顶部 返回列表