#### 原题说明

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

`max_L + pre_sum[i] - pre_sum[i - M]`表示以`i`结尾最后连续`M`个数，与之前最大的连续`L`个数的和。

#### 归纳总结

------ 关注公众号：猩猩的乐园 ------