题目如下：

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

解题思路：A的长度最大只有1000，表示O(n^2)的复杂度可以接受，那么直接用嵌套的两个循环暴力计算吧。

代码如下：

class Solution(object):
    def maxSumTwoNoOverlap(self, A, L, M):
        """
        :type A: List[int]
        :type L: int
        :type M: int
        :rtype: int
        """
        val = []
        count = 0
        for i in range(len(A)):
            count += A[i]
            val.append(count)

        res = 0
        for i in range(len(A)-L+1):
            for j in range(len(A)-M+1):
                if i == 0 and j == 2:
                    pass
                if (j+M-1) < i or (i+L-1) < j:
                    v2 = val[j+M-1]
                    if j>=1:
                        v2 -= val[j-1]
                    v1 = val[i+L-1]
                    if i >= 1:
                        v1 -= val[i - 1]
                    res = max(res, v1 + v2)
        return res

【leetcode】1031. Maximum Sum of Two Non-Overlapping Subarrays