1031. Maximum Sum of Two Non-Overlapping Subarrays

1、题目

Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M.  (For clarification, the L-length subarray could occur before or after the M-length subarray.)

Formally, return the largest V for which V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) and either:

• 0 <= i < i + L - 1 < j < j + M - 1 < A.length, or
• 0 <= j < j + M - 1 < i < i + L - 1 < A.length.

 

Example 1:

Input: A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.

Example 2:

Input: A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.

Example 3:

Input: A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3.

 

Note:

1. L >= 1
2. M >= 1
3. L + M <= A.length <= 1000
4. 0 <= A[i] <= 1000

2、代码

class Solution {
    public int maxSumTwoNoOverlap(int[] A, int L, int M) {
        if (A == null || A.length == 0){
            return 0;
        }
        int[] mediateMResult = new int[A.length - M + 1];
        for (int i = 0; i < M; i++){
            mediateMResult[0] += A[i];
        }
        
        for (int i = 1; i < mediateMResult.length; i++){
            mediateMResult[i] = mediateMResult[i - 1] - A[i - 1] + A[i + M - 1];   
        }
        
        int[] mediateLResult = new int[A.length - L + 1];
        for(int i = 0; i < L;i++){
            mediateLResult[0] += A[i];
        }
        for (int i = 1; i < mediateLResult.length; i++){
             mediateLResult[i] = mediateLResult[i - 1] - A[i - 1] + A[i + L - 1]; 
        }
        
        
        int result = 0;
        for (int i = 0; i < mediateMResult.length; i++){
            int firstValue = mediateMResult[i];
            int secondValue = Math.max(maxMediateSum(A, 0, i, mediateLResult, L),                            maxMediateSum(A, i + M, A.length, mediateLResult, L));
            result = Math.max(result, firstValue + secondValue);
        }
        return result;
    }
    

    public int maxMediateSum(int[] A, int begin, int end, int[] mediateResult, int L){
        if (end - begin < L){
            return -1;
        }
        int result = mediateResult[begin];
        for (int i = begin + 1; i <= end - L; i ++){
            result = Math.max(result, mediateResult[i]);
        }
        return result;
    }
}

 

优化版：

class Solution {
    public int maxSumTwoNoOverlap(int[] A, int L, int M) {
        if (A == null || A.length == 0){
            return 0;
        }
        
        for(int i = 1; i < A.length; i++){
            A[i] += A[i - 1];   
        }
        
        int Lmax = A[L -1];
        int Mmax = A[M - 1];
        int res = A[L + M - 1];
        for (int i = L + M ; i < A.length; i++){
            Lmax = Math.max(Lmax, A[i - M] - A[i - M - L]);
            Mmax = Math.max(Mmax, A[i - L] - A[i - L - M]);
            
            res = Math.max(res, Math.max(Lmax + A[i] - A[i - M], 
                                         Mmax + A[i] - A[i - L] ));
        }
        return res;
    }   
}