【leetcode】1031. Maximum Sum of Two Non-Overlapping Subarrays

题目如下：

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

解题思路：A的长度最大只有1000，表示O(n^2)的复杂度可以接受，那么直接用嵌套的两个循环暴力计算吧。

代码如下：

class Solution(object):

    def maxSumTwoNoOverlap(self, A, L, M):

        """

        :type A: List[int]

        :type L: int

        :type M: int

        :rtype: int

        """

        val = []

        count = 0

        for i in range(len(A)):

            count += A[i]

            val.append(count)

        res = 0

        for i in range(len(A)-L+1):

            for j in range(len(A)-M+1):

                if i == 0 and j == 2:

                    pass

                if (j+M-1) < i or (i+L-1) < j:

                    v2 = val[j+M-1]

                    if j>=1:

                        v2 -= val[j-1]

                    v1 = val[i+L-1]

                    if i >= 1:

                        v1 -= val[i - 1]

                    res = max(res, v1 + v2)

        return res

巴特西

【leetcode】1031. Maximum Sum of Two Non-Overlapping Subarrays

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