题目地址
https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/
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.
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
思路(动态规划)
题目中要求在前N(数组长度)个数中找出长度分别为L和M的非重叠子数组之和的最大值, 因此, 我们可以定义数组A中前i个数可构成的非重叠子数组L和M的最大值为SUMM[i], 并找到SUMM[i]和SUMM[i-1]的关系, 那么最终解就是SUMM[N]. 以下为图解:
关键点解析
- 注意图中描述的都是A[i-1], 而不是A[i], 因为base case为空数组, 而不是A[0];
- 求解图中ASUM数组的时候, 注意定义的是ASUM[i] = sum(A[0:i]), 因此当i等于0时, A[0:0]为空数组, 即: ASUM[0]为0, 而ASUM[1]才等于A[0];
- 求解图中MAXL数组时, 注意i < L时, 没有意义, 因为长度不够, 所以从i = L时才开始求解;
- 求解图中MAXM数组时, 也一样, 要从i = M时才开始求解;
- 求解图中SUMM数组时, 因为我们需要一个L子数组和一个M子数组, 因此长度要大于等于L+M才有意义, 所以要从i = L + M时开始求解.
代码
- 语言支持: Python
Python Code:
class Solution: def maxSumTwoNoOverlap(self, a: List[int], l: int, m: int) -> int: """ define asum[i] as the sum of subarray, a[0:i] define maxl[i] as the maximum sum of l-length subarray in a[0:i] define maxm[i] as the maximum sum of m-length subarray in a[0:i] define msum[i] as the maximum sum of non-overlap l-length subarray and m-length subarray case 1: a[i] is both not in l-length subarray and m-length subarray, then msum[i] = msum[i - 1] case 2: a[i] is in l-length subarray, then msum[i] = asum[i] - asum[i-l] + maxm[i-l] case 3: a[i] is in m-length subarray, then msum[i] = asum[i] - asum[i-m] + maxl[i-m] so, msum[i] = max(msum[i - 1], asum[i] - asum[i-l] + maxl[i-l], asum[i] - asum[i-m] + maxm[i-m]) """ alen, tlen = len(a), l + m asum = [0] * (alen + 1) maxl = [0] * (alen + 1) maxm = [0] * (alen + 1) msum = [0] * (alen + 1) for i in range(tlen): if i == 1: asum[i] = a[i - 1] elif i > 1: asum[i] = asum[i - 1] + a[i - 1] if i >= l: maxl[i] = max(maxl[i - 1], asum[i] - asum[i - l]) if i >= m: maxm[i] = max(maxm[i - 1], asum[i] - asum[i - m]) for i in range(tlen, alen + 1): asum[i] = asum[i - 1] + a[i - 1] suml = asum[i] - asum[i - l] summ = asum[i] - asum[i - m] maxl[i] = max(maxl[i - 1], suml) maxm[i] = max(maxm[i - 1], summ) msum[i] = max(msum[i - 1], suml + maxm[i - l], summ + maxl[i - m]) return msum[-1]
扩展
- 代码中, 求解了4个动态规划数组来求解最终值, 有没有可能只用两个数组来求解该题, 可以的话, 需要保留的又是哪两个数组?
- 代码中, 求解的4动态规划数组的顺序能否改变, 哪些能改, 哪些不能改?
如果采用前缀和数组的话，可以只使用O(n)的空间来存储前缀和，O(1)的动态规划状态空间来完成。C++代码如下:
class Solution { public: int maxSumTwoNoOverlap(vector<int>& A, int L, int M) { auto tmp = vector<int>{A[0]}; for (auto i = 1; i < A.size(); ++i) { tmp.push_back(A[i] + tmp[i - 1]); } auto res = tmp[L + M - 1], lMax = tmp[L - 1], mMax = tmp[M - 1]; for (auto i = L + M; i < tmp.size(); ++i) { lMax = max(lMax, tmp[i - M] - tmp[i - M - L]); mMax = max(mMax, tmp[i - L] - tmp[i - L - M]); res = max(res, max(lMax + tmp[i] - tmp[i - M], mMax + tmp[i] - tmp[i - L])); } return res; } };LeetCode题解1020.number-of-enclaves
题目地址 https://leetcode-cn.com/problems/number-of-enclaves/ 题目描述 给出一个二维数组 A，每个单元格为 0（代表海）或 1（代表陆地）。 移动是指在陆地上从一个地方走到另一个地方（朝四个方向之一）或离开网格的边界。 返回网格中无法在任意次数的移动中离开网格边界的陆地单元格的数量。 示例 1： 输入…