[LeetCode] 1031. Maximum Sum of Two Non-Overlapping Subarrays 兩個不重疊的子數組的最大和

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

這道題給了一個非負數組A，還有兩個長度L和M，說是要分別找出不重疊且長度分別為L和M的兩個子數組，前后順序無所謂，問兩個子數組最大的數字之和是多少。博主最開始想的方法是用動態規划 Dynamic Programming 和滑動窗口 Sliding Window 來做，用兩個 dp 數組，其中 front[i] 表示范圍 [0, i] 之間的長度為M的子數組的最大數字之和，back[i] 表示范圍 [i, n-1] 之間的長度為M的子數組的最大數字之和。然后再次遍歷數組，維護一個長度為L的滑動數組，當數組長度正好為L的時候，當前窗口的數字之和加上左邊的 front[left-1]，或者加上右邊的 back[i+1]，取二者中的較大值來更新結果 res，這種解法可以通過 OJ，但是行數比較多，且用了三個 for 循環，這里就不貼了。來看論壇上的高分解法吧，首先建立累加和數組，這里可以直接覆蓋A數組，然后定義 Lmax 為在最后M個數字之前的長度為L的子數組的最大數字之和，同理，Mmax 表示在最后L個數字之前的長度為M的子數組的最大數字之和。結果 res 初始化為前 L+M 個數字之和，然后遍歷數組，從 L+M 開始遍歷，先更新 Lmax 和 Mmax，其中 Lmax 用 A[i - M] - A[i - M - L] 來更新，Mmax 用 A[i - L] - A[i - M - L] 來更新。然后取 Lmax + A[i] - A[i - M] 和 Mmax + A[i] - A[i - L] 之間的較大值來更新結果 res 即可，參見代碼如下：

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

參考資料：

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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