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
, or0 <= 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
给出非负整数数组
A
,返回两个非重叠(连续)子数组中元素的最大和,子数组的长度分别为 L
和 M
。(这里需要澄清的是,长为 L 的子数组可以出现在长为 M 的子数组之前或之后。)
从形式上看,返回最大的 V
,而 V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1])
并满足下列条件之一:
0 <= i < i + L - 1 < j < j + M - 1 < A.length
, 或0 <= j < j + M - 1 < i < i + L - 1 < A.length
.
示例 1:
输入:A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2 输出:20 解释:子数组的一种选择中,[9] 长度为 1,[6,5] 长度为 2。
示例 2:
输入:A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2 输出:29 解释:子数组的一种选择中,[3,8,1] 长度为 3,[8,9] 长度为 2。
示例 3:
输入:A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3 输出:31 解释:子数组的一种选择中,[5,6,0,9] 长度为 4,[0,3,8] 长度为 3。
提示:
L >= 1
M >= 1
L + M <= A.length <= 1000
0 <= A[i] <= 1000
Runtime: 24 ms
Memory Usage: 19.1 MB
1 class Solution { 2 let N:Int = 1010 3 var prefix:[Int] = [Int](repeating:0,count:1010) 4 func maxSumTwoNoOverlap(_ A: [Int], _ L: Int, _ M: Int) -> Int { 5 var n:Int = A.count 6 for i in 1...n 7 { 8 prefix[i] = prefix[i - 1] + A[i - 1] 9 } 10 var ans:Int = 0 11 var best_l:Int = 0 12 for i in (L + M)...n 13 { 14 best_l = max(best_l, prefix[i - M] - prefix[i - M - L]) 15 ans = max(ans, best_l + prefix[i] - prefix[i - M]) 16 } 17 var best_m:Int = 0 18 for i in (L + M)...n 19 { 20 best_m = max(best_m, prefix[i - L] - prefix[i - M - L]) 21 ans = max(ans, best_m + prefix[i] - prefix[i - L]) 22 } 23 return ans 24 } 25 }
原文地址:https://www.cnblogs.com/strengthen/p/10744670.html
时间: 2024-10-07 21:43:31