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

2021-02-15 15:05:38 阅读：243 来源： 互联网

标签：Mmax Non res Subarrays Sum length Lmax 数组 subarrays

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

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] 和 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

LeetCode All in One 题目讲解汇总(持续更新中...)

标签：Mmax,Non,res,Subarrays,Sum,length,Lmax,数组,subarrays
来源： https://www.cnblogs.com/grandyang/p/14403710.html

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