689 Maximum Sum of 3 Non-Overlapping Subarrays 三个无重叠子数组的最大和
Description:
Given an integer array nums and an integer k, find three non-overlapping subarrays of length k with maximum sum and return them.
Return the result as a list of indices representing the starting position of each interval (0-indexed). If there are multiple answers, return the lexicographically smallest one.
Example:
Example 1:
Input: nums = [1,2,1,2,6,7,5,1], k = 2
Output: [0,3,5]
Explanation: Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5].
We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger.
Example 2:
Input: nums = [1,2,1,2,1,2,1,2,1], k = 2
Output: [0,2,4]
Constraints:
1 <= nums.length <= 2 * 10^4
1 <= nums[i] < 2^16
1 <= k <= floor(nums.length / 3)
题目描述:
给定数组 nums 由正整数组成，找到三个互不重叠的子数组的最大和。
每个子数组的长度为k，我们要使这3*k个项的和最大化。
返回每个区间起始索引的列表（索引从 0 开始）。如果有多个结果，返回字典序最小的一个。
示例 :
输入: [1,2,1,2,6,7,5,1], 2
输出: [0, 3, 5]
解释: 子数组 [1, 2], [2, 6], [7, 5] 对应的起始索引为 [0, 3, 5]。
我们也可以取 [2, 1], 但是结果 [1, 3, 5] 在字典序上更大。
注意:
nums.length的范围在[1, 20000]之间。
nums[i]的范围在[1, 65535]之间。
k的范围在[1, floor(nums.length / 3)]之间。
思路:
动态规划
用一个 window 数组记录所有长度为 k 的子数组的和
用两个数组 left, right 记录 window[i] 的最大值对应的下标
其中 left 数组记录 [0, i] 中 window[i] 的最大值对应的下标
right 数组记录 [i, window.size() - 1] 的最大值对应的下标
最后再遍历所有的可能值, 找到中间下标 i 使得 window[left[i - k]] + window[i] + window[right[i + k]] 最大, 返回对应的 { left[i - k], k, right[i + k] } 数组
时间复杂度为 O(n), 空间复杂度为 O(n)
代码:
C++:
class Solution
{
public:
vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k)
{
int n = nums.size(), cur = accumulate(nums.begin(), nums.begin() + k, 0);
vector<int> window(n - k + 1, cur), left(n - k + 1), right(n - k + 1, n - k), result;
for (int i = k; i < n; i++) window[i - k + 1] = window[i - k] + nums[i] - nums[i - k];
for (int i = 1; i < n - k + 1; i++)
{
if (window[i] > window[left[i - 1]]) left[i] = i;
else left[i] = left[i - 1];
if (window[n - k - i] >= window[right[n - k - i + 1]]) right[n - k - i] = n - k - i;
else right[n - k - i] = right[n - k - i + 1];
}
cur = 0;
for (int i = k; i < n - (k << 1) + 1; i++)
{
if (cur < window[i] + window[left[i - k]] + window[right[i + k]])
{
cur = window[i] + window[left[i - k]] + window[right[i + k]];
result = vector<int>{ left[i - k], i, right[i + k] };
}
}
return result;
}
};
Java:
class Solution {
public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
int n = nums.length, cur = 0, result[] = new int[3];
int window[] = new int[n - k + 1], left[] = new int[n - k + 1], right[] = new int[n - k + 1];
for (int i = 0; i < k; i++) cur += nums[i];
Arrays.fill(window, cur);
for (int i = k; i < n; i++) window[i - k + 1] = window[i - k] + nums[i] - nums[i - k];
Arrays.fill(right, n - k);
for (int i = 1; i < n - k + 1; i++) {
if (window[i] > window[left[i - 1]]) left[i] = i;
else left[i] = left[i - 1];
if (window[n - k - i] >= window[right[n - k - i + 1]]) right[n - k - i] = n - k - i;
else right[n - k - i] = right[n - k - i + 1];
}
cur = 0;
for (int i = k; i < n - (k << 1) + 1; i++) {
if (cur < window[i] + window[left[i - k]] + window[right[i + k]]) {
cur = window[i] + window[left[i - k]] + window[right[i + k]];
result = new int[]{ left[i - k], i, right[i + k] };
}
}
return result;
}
}
Python:
class Solution:
def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
n = len(nums)
window, left, right, cur, result = [sum(nums[:k])] * (n - k + 1), [0] * (n - k + 1), [n - k] * (n - k + 1), 0, [0] * 3
for i in range(k, n):
window[i - k + 1] = window[i - k] - nums[i - k] + nums[i]
for i in range(1, n - k + 1):
if window[i] > window[left[i - 1]]:
left[i] = i
else:
left[i] = left[i - 1]
if window[n - k - i] >= window[right[n - k - i + 1]]:
right[n - k - i] = n - k - i
else:
right[n - k - i] = right[n - k - i + 1]
for i in range(k, n - k + 1 - k):
if cur < window[i] + window[left[i - k]] + window[right[i + k]]:
cur = window[i] + window[left[i - k]] + window[right[i + k]]
result = [left[i - k], i, right[i + k]]
return result
