LeetCode 1043. Partition Array for Maximum Sum

佚名 6年前 (2019-05-12) 随笔 515人围观抢沙发百度已收录

Problem description:

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.

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Return the largest sum of the given array after partitioning.

题解：dp

for(j = i; j > i - K && j >= 0; j--) {
                maxInPart = Math.max(maxInPart, A[j]);
                int prev = 0;
                if(j - 1 >= 0) prev = dp[j - 1];
                dp[i] = Math.max(dp[i], prev + maxInPart * (i - j + 1));
}

时间复杂度:O(kn), 空间复杂度:O(n)

class Solution {
    public int maxSumAfterPartitioning(int[] A, int K) {
        int n = A.length;
        int[] dp = new int[n];
        dp[0] = A[0];
        for(int i = 1; i < n; i++) {
            int j = i;
            int maxInPart = A[j];
            for(j = i; j > i - K && j >= 0; j--) {
                maxInPart = Math.max(maxInPart, A[j]);
                int prev = 0;
                if(j - 1 >= 0) prev = dp[j - 1];
                dp[i] = Math.max(dp[i], prev + maxInPart * (i - j + 1));
                
            }
            
        }
        
        return dp[n - 1];
    }
}