Leetcode 1031 Maximum Sum of Two Non-Overlapping Subarrays (滑动窗口) _overlapping

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Leetcode 1031
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.

例子

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.

限制条件

Note: L > = 1 M > = 1 L + M < = A.length < = 1000 0 < = A[i] < = 1000

解题思路
? 记A[i]为A[0:i+1]的和；
? 记Lmax为除末尾M个元素外的最大L长度子序列的和；
? 记Mmax为除末尾N个元素外的最大M长度子序列的和；
? 记res为当前L子序列、M子序列和的最大值；
? 遍历A[L+M:A.length], 求res, ( Lmax+末尾M个元素 ), ( Mmax+末尾L个元素 ) 中的最大值。
【Leetcode 1031 Maximum Sum of Two Non-Overlapping Subarrays (滑动窗口)】? 由题目规模较小，暴力算法也可解决该问题。
代码

python3 class Solution: def maxSumTwoNoOverlap(self, A: List[int], L: int, M: int) -> int: for i in range(1,len(A)): A[i] += A[i-1] Lmax, Mmax, res = A[L-1], A[M-1], A[L+M-1] for i in range(L+M, len(A)): Lmax = max(Lmax, A[i-M]-A[i-M-L]) Mmax = max(Mmax, A[i-L]-A[i-M-L]) res = max(res, Lmax+A[i]-A[i-M], Mmax+A[i]-A[i-L]) return res

java class Solution { public int maxSumTwoNoOverlap(int[] A, int L, int M) { for(int i=1; i< A.length; ++i){ A[i] += A[i-1]; } int Lmax=A[L-1], Mmax=A[M-1], res=A[L+M-1]; for(int i=L+M; i< A.length; ++i){ Lmax = Math.max(Lmax, A[i-M]-A[i-M-L]); Mmax = Math.max(Mmax, A[i-L]-A[i-M-L]); res = Math.max(res, Math.max(Lmax+A[i]-A[i-M], Mmax+A[i]-A[i-L])); } return res; } }