1031. Maximum Sum of Two Non-Overlapping Subarrays**

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珍妮的选择 2022-05-30 11:08:38 博主文章分类：LeetCode ©著作权

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1031. Maximum Sum of Two Non-Overlapping Subarrays**

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

C++ 实现 1

思路来自: [Java/C++/Python] O(N)Time O(1) Space

1031. Maximum Sum of Two Non-Overlapping Subarrays**_算法

其实上面都是废话, 思路都在代码里.

注意先对 A[0, ... i] 子序列求和, 然后注意 res, maxL, maxM 的初始化.

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]; // A[i] = sum(A[0 ... i])
        int res = A[L + M - 1], maxL = A[L - 1], maxM = A[M - 1];
        for (int i = L + M; i < A.size(); ++ i) {
            maxL = std::max(maxL, A[i - M] - A[i - M - L]);
            maxM = std::max(maxM, A[i - L] - A[i - M - L]);
            res = std::max(res, std::max(maxL + A[i] - A[i - M], maxM + A[i] - A[i - L]));
        }
        return res;
    }
};