leetcode1043. Partition Array for Maximum Sum

Posted 2022-01-20 seyjs

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. 1 <= K <= A.length <= 500
2. 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]