#1031 动态规划: 求解数组中两个非重叠子数组的最大和, O(n)算法 (azl397985856#25)

README.en.md

@@ -181,6 +181,7 @@ The data structures mainly includes:
 - [0877.stone-game](./problems/877.stone-game.md)
 - [0887.super-egg-drop](./problems/887.super-egg-drop.md)
 - [0900.rle-iterator](./problems/900.rle-iterator.md)
+- [1031.maximum-sum-of-two-non-overlapping-subarrays](./problems/1031.maximum-sum-of-two-non-overlapping-subarrays.md) 🆕
 
 #### Hard
 - [0023.merge-k-sorted-lists](./problems/23.merge-k-sorted-lists.md)

README.md

@@ -179,6 +179,7 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - [0877.stone-game](./problems/877.stone-game.md)
 - [0887.super-egg-drop](./problems/887.super-egg-drop.md)
 - [0900.rle-iterator](./problems/900.rle-iterator.md)
+- [1031.maximum-sum-of-two-non-overlapping-subarrays](./problems/1031.maximum-sum-of-two-non-overlapping-subarrays.md) 🆕
 
 #### 困难难度
 - [0023.merge-k-sorted-lists](./problems/23.merge-k-sorted-lists.md)

assets/problems/1031.maximum-sum-of-two-non-overlapping-subarrays.drawio

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problems/1031.maximum-sum-of-two-non-overlapping-subarrays.md

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+## 题目地址
+https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/
+
+## 题目描述
+
+```
+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:
+
+L >= 1
+M >= 1
+L + M <= A.length <= 1000
+0 <= A[i] <= 1000
+```
+
+## 思路(动态规划)
+
+题目中要求在前N(数组长度)个数中找出长度分别为L和M的非重叠子数组之和的最大值, 因此, 我们可以定义数组A中前i个数可构成的非重叠子数组L和M的最大值为SUMM[i], 并找到SUMM[i]和SUMM[i-1]的关系, 那么最终解就是SUMM[N]. 以下为图解:
+
+![1031.Maximum Sum of Two Non-Overlapping Subarrays](../assets/problems/1031.maximum-sum-of-two-non-overlapping-subarrays.png)
+
+## 关键点解析
+
+1. 注意图中描述的都是A[i-1], 而不是A[i], 因为base case为空数组, 而不是A[0];
+2. 求解图中ASUM数组的时候, 注意定义的是ASUM[i] = sum(A[0:i]), 因此当i等于0时, A[0:0]为空数组, 即: ASUM[0]为0, 而ASUM[1]才等于A[0];
+3. 求解图中MAXL数组时, 注意i < L时, 没有意义, 因为长度不够, 所以从i = L时才开始求解;
+4. 求解图中MAXM数组时, 也一样, 要从i = M时才开始求解;
+5. 求解图中SUMM数组时, 因为我们需要一个L子数组和一个M子数组, 因此长度要大于等于L+M才有意义, 所以要从i = L + M时开始求解.
+
+## 代码
+
+* 语言支持: Python
+
+Python Code:
+```python
+class Solution:
+    def maxSumTwoNoOverlap(self, a: List[int], l: int, m: int) -> int:
+        """
+
+        define asum[i] as the sum of subarray, a[0:i]
+        define maxl[i] as the maximum sum of l-length subarray in a[0:i]
+        define maxm[i] as the maximum sum of m-length subarray in a[0:i]
+        define msum[i] as the maximum sum of non-overlap l-length subarray and m-length subarray
+
+        case 1: a[i] is both not in l-length subarray and m-length subarray, then msum[i] = msum[i - 1]
+        case 2: a[i] is in l-length subarray, then msum[i] = asum[i] - asum[i-l] + maxm[i-l]
+        case 3: a[i] is in m-length subarray, then msum[i] = asum[i] - asum[i-m] + maxl[i-m]
+
+        so, msum[i] = max(msum[i - 1], asum[i] - asum[i-l] + maxl[i-l], asum[i] - asum[i-m] + maxm[i-m])
+        """
+
+        alen, tlen = len(a), l + m
+        asum = [0] * (alen + 1)
+        maxl = [0] * (alen + 1)
+        maxm = [0] * (alen + 1)
+        msum = [0] * (alen + 1)
+
+        for i in range(tlen):
+            if i == 1:
+                asum[i] = a[i - 1]
+            elif i > 1:
+                asum[i] = asum[i - 1] + a[i - 1]
+            if i >= l:
+                maxl[i] = max(maxl[i - 1], asum[i] - asum[i - l])
+            if i >= m:
+                maxm[i] = max(maxm[i - 1], asum[i] - asum[i - m])
+
+        for i in range(tlen, alen + 1):
+            asum[i] = asum[i - 1] + a[i - 1]
+            suml = asum[i] - asum[i - l]
+            summ = asum[i] - asum[i - m]
+            maxl[i] = max(maxl[i - 1], suml)
+            maxm[i] = max(maxm[i - 1], summ)
+            msum[i] = max(msum[i - 1], suml + maxm[i - l], summ + maxl[i - m])
+
+        return msum[-1]
+```
+
+## 扩展
+
+1. 代码中, 求解了4个动态规划数组来求解最终值, 有没有可能只用两个数组来求解该题, 可以的话, 需要保留的又是哪两个数组?
+2. 代码中, 求解的4动态规划数组的顺序能否改变, 哪些能改, 哪些不能改?