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
给出整数数组 A
,将该数组分隔为长度最多为 K 的几个(连续)子数组。分隔完成后,每个子数组的中的值都会变为该子数组中的最大值。
返回给定数组完成分隔后的最大和。
示例:
输入:A = [1,15,7,9,2,5,10], K = 3 输出:84 解释:A 变为 [15,15,15,9,10,10,10]
提示:
1 <= K <= A.length <= 500
0 <= A[i] <= 10^6
Runtime: 56 ms
Memory Usage: 20.9 MB
1 class Solution { 2 func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int { 3 let n:Int = A.count 4 var dp:[Int] = [Int](repeating:0,count:n + 1) 5 for i in 1...n 6 { 7 var maxs:Int = 0 8 var j:Int = 1 9 while(j <= K && i - j >= 0) 10 { 11 maxs = max(maxs, A[i - j]) 12 dp[i] = max(dp[i], dp[i - j] + maxs * j) 13 j += 1 14 } 15 } 16 return dp[n] 17 } 18 }