[LeetCode] 1031. Maximum Sum of Two Non-Overlapping Subarrays 两个不重叠的子数组的最大和
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
这道题给了一个非负数组A,还有两个长度L和M,说是要分别找出不重叠且长度分别为L和M的两个子数组,前后顺序无所谓,问两个子数组最大的数字之和是多少。博主最开始想的方法是用动态规划 Dynamic Programming 和滑动窗口 Sliding Window 来做,用两个 dp 数组,其中 front[i] 表示范围 [0, i] 之间的长度为M的子数组的最大数字之和,back[i] 表示范围 [i, n-1] 之间的长度为M的子数组的最大数字之和。然后再次遍历数组,维护一个长度为L的滑动数组,当数组长度正好为L的时候,当前窗口的数字之和加上左边的 front[left-1],或者加上右边的 back[i+1],取二者中的较大值来更新结果 res,这种解法可以通过 OJ,但是行数比较多,且用了三个 for 循环,这里就不贴了。来看论坛上的高分解法吧,首先建立累加和数组,这里可以直接覆盖A数组,然后定义 Lmax 为在最后M个数字之前的长度为L的子数组的最大数字之和,同理,Mmax 表示在最后L个数字之前的长度为M的子数组的最大数字之和。结果 res 初始化为前 L+M 个数字之和,然后遍历数组,从 L+M 开始遍历,先更新 Lmax 和 Mmax,其中 Lmax 用 A[i - M] - A[i - M - L]
来更新,Mmax 用 A[i - L] - A[i - M - L]
来更新。然后取 Lmax + A[i] - A[i - M]
和 Mmax + A[i] - A[i - L]
之间的较大值来更新结果 res 即可,参见代码如下:
class Solution {
public:
int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
for (int i = 1; i < A.size(); ++i) {
A[i] += A[i - 1];
}
int res = A[L + M - 1], Lmax = A[L - 1], Mmax = A[M - 1];
for (int i = L + M; i < A.size(); ++i) {
Lmax = max(Lmax, A[i - M] - A[i - M - L]);
Mmax = max(Mmax, A[i - L] - A[i - M - L]);
res = max(res, max(Lmax + A[i] - A[i - M], Mmax + A[i] - A[i - L]));
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1031
参考资料:
https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/