feat: #1031: add concise C++ implementation (#164)

Author: raof01

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

@@ -104,3 +104,23 @@ class Solution:
 
 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?
+
+如果采用前缀和数组的话，可以只使用O(n)的空间来存储前缀和，O(1)的动态规划状态空间来完成。C++代码如下:
+```C++
+class Solution {
+public:
+    int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
+        auto tmp = vector<int>{A[0]};
+        for (auto i = 1; i < A.size(); ++i) {
+            tmp.push_back(A[i] + tmp[i - 1]);
+        }
+        auto res = tmp[L + M - 1], lMax = tmp[L - 1], mMax = tmp[M - 1];
+        for (auto i = L + M; i < tmp.size(); ++i) {
+            lMax = max(lMax, tmp[i - M] - tmp[i - M - L]);
+            mMax = max(mMax, tmp[i - L] - tmp[i - L - M]);
+            res = max(res, max(lMax + tmp[i] - tmp[i - M], mMax + tmp[i] - tmp[i - L]));
+        }
+        return res;
+    }
+};
+```