【leetcode】1043. Partition Array for Maximum Sum

題目如下：

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

解題思路：假設dp[i][j] 表示第i個元素爲第j個子數組的最後一個元素時，A[0:i]可以獲得的最大值。那麼有dp[i][j] = max(dp[i][j], dp[m][j-1] + max_val[m+1][i] * (i-m))   ( i-k < m < i) 。

代碼如下：

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]