【leetcode】1043. Partition Array for Maximum Sum

最新推荐文章于 2023-06-13 20:11:05 发布

weixin_34281477

最新推荐文章于 2023-06-13 20:11:05 发布

阅读量190

点赞数

CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/seyjs/p/11044749.html

博客给出一道数组分区求最大和的题目，即给定整数数组，将其划分为长度至多为k的连续子数组，分区后各子数组值变为其最大值，求分区后数组最大和。还给出解题思路，用dp[i][j]表示相关最大值，并给出状态转移方程，最后给出代码来源。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目如下：

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

解题思路：假设dp[i][j] 表示第i个元素为第j个子数组的最后一个元素时，A[0:i]可以获得的最大值。那么有dp[i][j] = max(dp[i][j], dp[m][j-1] + max_val[m+1][i] * (i-m)) ( i-k < m < i) 。

代码如下：

class Solution(object):
    def maxSumAfterPartitioning(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: int
        """
        import math
        dp = []
        max_val = []
        sub = int(math.ceil(float(len(A))/K))
        for i in A:
            dp.append([0] * sub)
            max_val.append([0]*len(A))
        for i in range(len(A)):
            max_val[i][i] = A[i]
            for j in range(i+1,len(A)):
                max_val[i][j] = max(max_val[i][j-1],A[j])
        dp[0][0] = A[0]
        for i in range(len(A)):
            for j in range(sub):
                #print i,j
                if i-K< 0:
                    dp[i][j] = max(A[0:i+1]) * (i+1)
                else:
                    for m in range(i-K,i):
                        dp[i][j] = max(dp[i][j], dp[m][j-1] + max_val[m+1][i] * (i-m))
        #print dp
        return dp[-1][-1]