LeetCode 1043. Partition Array for Maximum Sum

原题链接在这里：https://leetcode.com/problems/partition-array-for-maximum-sum/

题目：

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

题解：

When encounter such kind of problem.

Could think from a simpler example. Say only one element, then 2 elements and more. Virtualize them, find routines.

Use array dp to memorize maxmimum sum up to i.

If, A = [1], then A becomes A[1]. dp = [1].

A = [1, 15], then A becomes A[15, 15]. dp = [1, 30].

A = [1, 15, 7], then A becomes A[15, 15, 15]. dp = [1, 30, 45].

A = [1, 15, 7, 9], then A becomes A[15, 15, 15, 9]. dp = [1, 30, 45, 54].

...

The routine is like from i back k(<= K) steps, find the maxmimum element, curMax * k + dp[i-k](if available).

Finally return dp[A.length-1].

Time Complexity: O(n*K). n = A.length.

Space: O(n).

AC Java:

 class Solution {

     public int maxSumAfterPartitioning(int[] A, int K) {

         int len = A.length;

         int [] dp = new int[len];

         for(int i = 0; i<len; i++){

             dp[i] = Integer.MIN_VALUE;

             int curMax = A[i];

             for(int k = 1; k<=K & i-k+1>=0; k++){

                 curMax = Math.max(curMax, A[i-k+1]);

                 dp[i] = Math.max(dp[i], (i-k<0 ? 0 : dp[i-k]) + curMax*k);

             }

         }

         return dp[len-1];

     }

 }

巴特西

LeetCode 1043. Partition Array for Maximum Sum

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