Leetcode 2874. Maximum Value of an Ordered Triplet II

Espresso Macchiato2023-10-04 17:24

[Leetcode 2874. Maximum Value of an Ordered Triplet II](#Leetcode 2874. Maximum Value of an Ordered Triplet II)
- [1. 解题思路](#1. 解题思路)
- [2. 代码实现](#2. 代码实现)

题目链接：Leetcode 2874. Maximum Value of an Ordered Triplet II

1. 解题思路

这一题我的思路的核心点在于将中间点j作为关键点，此时，要令triplet最大，事实上就是分别在其左右找到最大元素即可，而这个就是一个类似累计数组的算法，倒是无需赘述了......

2. 代码实现

给出python代码实现如下：

python 复制代码

class Solution:
    def maximumTripletValue(self, nums: List[int]) -> int:
        n = len(nums)
        left, right = deepcopy(nums), deepcopy(nums)
        
        for i in range(n-1):
            left[i+1] = max(left[i], left[i+1])
            
        for i in range(n-2, -1, -1):
            right[i] = max(right[i+1], right[i])
            
        res = max((left[i-1] - nums[i]) * right[i+1] for i in range(1, n-1))
        return max(res, 0)

提交代码评测得到：耗时1185ms，占用内存29MB。