[leetcode] 1031. Maximum Sum of Two Non-Overlapping Subarrays

原创

是念 2022-08-12 07:17:23 博主文章分类：python ©著作权

©著作权归作者所有：来自51CTO博客作者是念的原创作品，请联系作者获取转载授权，否则将追究法律责任

Description

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

分析

题目的意思是：给你一个数组，求取出L长度的子数组和M长度的子数组，两数组不相交，这道题我不会，找了一个我能看懂的解法，首先是建立一个求和的数组，然后再构造题目中说的两种情况，找出最大值就行了，注意两次遍历的取的遍历区间，特别是n+1,n+1代表的是prefix数组的长度哈，因为最开始加了一个0哈。

代码

class Solution:
    def maxSumTwoNoOverlap(self, A: List[int], L: int, M: int) -> int:
        n=len(A)
        prefix=[0]
        for i in range(n):
            t=prefix[-1]+A[i]
            prefix.append(t)
        maxl=maxm=0
        summ=0
        for i in range(M,n+1-L):
            maxm=max(maxm,prefix[i]-prefix[i-M])
            summ=max(summ,maxm+prefix[i+L]-prefix[i])
            
        for i in range(L,n+1-M):
            maxl=max(maxl,prefix[i]-prefix[i-L])
            summ=max(summ,maxl+prefix[i+M]-prefix[i])
        return summ