LeetCode1031.不重叠连续子数组最大和

1031. Maximum Sum of Two Non-Overlapping 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:

1. L >= 1
2. M >= 1
3. L + M <= A.length <= 1000
4. 0 <= A[i] <= 1000

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题目：数组A中不重叠的两个连续子数组（长度分别为L和M）的最大和。

思路：参见Solution。当子串L在M左侧时，用l_max表示A[0...i](i+M)中子串L的最大和，两个子数组的最大和可以表示为l_max + sum(A[i:i+M])；当L在M右边时时，用m_max表示A[0...i](i+L)中子串M的最大和，两个子数组的最大和为m_max + sum(A[i:i+L])。取两种情况的较大值。

工程代码下载

class Solution {
public:
    int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
        int n = A.size();
        if (n == 0)
            return 0;
        vector<int> sum(n);
        sum[0] = A[0];
        for (int i = 1; i < n; ++i) {
            sum[i] = sum[i-1] + A[i];
        }

        int res = sum[L + M - 1];
        int l_max = sum[L - 1];
        for (int i = L; i < n - M; ++i) {
            l_max = max(l_max, sum[i] - sum[i - L]);
            res = max(res, l_max + sum[i + M] - sum[i]);
        }

        int m_max = sum[M - 1];
        for (int i = M; i < n - L; ++i) {
            m_max = max(m_max, sum[i] - sum[i - M]);
            res = max(res, m_max + sum[i + L] - sum[i]);
        }

        return res;
    }
};