1043. Partition Array for Maximum Sum

最新推荐文章于 2023-06-13 20:11:05 发布

原创最新推荐文章于 2023-06-13 20:11:05 发布 · 285 阅读

1 ·

CC 4.0 BY-SA版权

leetcode 同时被 2 个专栏收录

558 篇文章

订阅专栏

87 篇文章

订阅专栏

本文介绍了一种基于动态规划(DP)的算法，用于解决给定数组A和整数K的最大分区求和问题。通过将数组分割成长度不超过K的连续子数组，并将每个子数组的元素替换为该子数组的最大值，目标是找到这些子数组的最大总和。文章提供了Python实现代码，包括处理边界条件和计算最优解的过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an integer array A, you partition the array into (contiguous) subarrays of length at most K. After partitioning, each subarray has their values changed to become the maximum value of that subarray.

Return the largest sum of the given array after partitioning.

Example 1:

Input: A = [1,15,7,9,2,5,10], K = 3
Output: 84
Explanation: A becomes [15,15,15,9,10,10,10]

Note:

1 <= K <= A.length <= 500
0 <= A[i] <= 10^6

思路：DP，

class Solution(object):
    def maxSumAfterPartitioning(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: int
        """
        n=len(A)
        ma=[[0 for _ in range(n)] for _ in range(n)]
        for i in range(n):
            ma[i][i]=A[i]
            for j in range(i+1,n): ma[i][j]=max(ma[i][j-1],A[j])
        
        dp=[0 for _ in range(n)]
        dp[0]=A[0]
        for i in range(1,n):
            for j in range(1,min(K+1,i+2)):
                if i-j<0: dp[i]=max(dp[i],ma[0][i]*(i+1))
                else: dp[i]=max(dp[i],dp[i-j]+ma[i-j+1][i]*j)
        
        return dp[-1]

做前面那题做崩了，一开始连题目也看错了，各种边界也没处理好