Leetcode 1031 Maximum Sum of Two Non-Overlapping Subarrays （两个不重叠子数组的最大和) (滑动窗口)

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。

  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;
    }
}