• # [LeetCode] 1031. Maximum Sum of Two Non-Overlapping Subarrays 两个不重叠的子数组的最大和

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`

这道题给了一个非负数组A，还有两个长度L和M，说是要分别找出不重叠且长度分别为L和M的两个子数组，前后顺序无所谓，问两个子数组最大的数字之和是多少。博主最开始想的方法是用动态规划 Dynamic Programming 和滑动窗口 Sliding Window 来做，用两个 dp 数组，其中 front[i] 表示范围 [0, i] 之间的长度为M的子数组的最大数字之和，back[i] 表示范围 [i, n-1] 之间的长度为M的子数组的最大数字之和。然后再次遍历数组，维护一个长度为L的滑动数组，当数组长度正好为L的时候，当前窗口的数字之和加上左边的 front[left-1]，或者加上右边的 back[i+1]，取二者中的较大值来更新结果 res，这种解法可以通过 OJ，但是行数比较多，且用了三个 for 循环，这里就不贴了。来看论坛上的高分解法吧，首先建立累加和数组，这里可以直接覆盖A数组，然后定义 Lmax 为在最后M个数字之前的长度为L的子数组的最大数字之和，同理，Mmax 表示在最后L个数字之前的长度为M的子数组的最大数字之和。结果 res 初始化为前 L+M 个数字之和，然后遍历数组，从 L+M 开始遍历，先更新 Lmax 和 Mmax，其中 Lmax 用 `A[i - M] - A[i - M - L]` 来更新，Mmax 用 `A[i - L] - A[i - M - L]` 来更新。然后取 `Lmax + A[i] - A[i - M]``Mmax + A[i] - A[i - L]` 之间的较大值来更新结果 res 即可，参见代码如下：

``````class Solution {
public:
int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
for (int i = 1; i < A.size(); ++i) {
A[i] += A[i - 1];
}
int res = A[L + M - 1], Lmax = A[L - 1], Mmax = A[M - 1];
for (int i = L + M; i < A.size(); ++i) {
Lmax = max(Lmax, A[i - M] - A[i - M - L]);
Mmax = max(Mmax, A[i - L] - A[i - M - L]);
res = max(res, max(Lmax + A[i] - A[i - M], Mmax + A[i] - A[i - L]));
}
return res;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1031

参考资料：

https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/

https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/discuss/278251/JavaC%2B%2BPython-O(N)Time-O(1)-Space

https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/discuss/279221/JavaPython-3-two-easy-DP-codes-w-comment-time-O(n)-NO-change-of-input

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• 原文地址：https://www.cnblogs.com/grandyang/p/14403710.html