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【每日一题】689. 三个无重叠子数组的最大和-2023.11.19

news 2025/8/28 6:47:37

题目：

689. 三个无重叠子数组的最大和

给你一个整数数组 nums 和一个整数 k ，找出三个长度为 k 、互不重叠、且全部数字和（3 * k 项）最大的子数组，并返回这三个子数组。

以下标的数组形式返回结果，数组中的每一项分别指示每个子数组的起始位置（下标从 0 开始）。如果有多个结果，返回字典序最小的一个。

示例 1：

输入：nums = [1,2,1,2,6,7,5,1], k = 2
输出：[0,3,5]
解释：子数组 [1, 2], [2, 6], [7, 5] 对应的起始下标为 [0, 3, 5]。
也可以取 [2, 1], 但是结果 [1, 3, 5] 在字典序上更大。

示例 2：

输入：nums = [1,2,1,2,1,2,1,2,1], k = 2
输出：[0,2,4]

提示：

1 <= nums.length <= 2 * 104
1 <= nums[i] < 216
1 <= k <= floor(nums.length / 3)

解答：

class Solution {public int[] maxSumOfOneSubarray(int[] nums, int k) {int[] ans = new int[1];int sum1 = 0, maxSum1 = 0;for (int i = 0; i < nums.length; ++i) {sum1 += nums[i];if (i >= k - 1) {if (sum1 > maxSum1) {maxSum1 = sum1;ans[0] = i - k + 1;}sum1 -= nums[i - k + 1];}}return ans;}
}

class Solution {public int[] maxSumOfTwoSubarrays(int[] nums, int k) {int[] ans = new int[2];int sum1 = 0, maxSum1 = 0, maxSum1Idx = 0;int sum2 = 0, maxSum12 = 0;for (int i = k; i < nums.length; ++i) {sum1 += nums[i - k];sum2 += nums[i];if (i >= k * 2 - 1) {if (sum1 > maxSum1) {maxSum1 = sum1;maxSum1Idx = i - k * 2 + 1;}if (maxSum1 + sum2 > maxSum12) {maxSum12 = maxSum1 + sum2;ans[0] = maxSum1Idx;ans[1] = i - k + 1;}sum1 -= nums[i - k * 2 + 1];sum2 -= nums[i - k + 1];}}return ans;}
}

代码：

class Solution {public int[] maxSumOfThreeSubarrays(int[] nums, int k) {int[] ans=new int[3];int sum1=0,maxSum1=0,maxSum1Idx=0;int sum2=0,maxSum12=0,maxSum12Idx1=0,maxSum12Idx2=0;int sum3=0,maxTotal=0;for(int i=2*k;i<nums.length;i++){sum1+=nums[i-2*k];sum2+=nums[i-k];sum3+=nums[i];if(i>=3*k-1){if(sum1>maxSum1){maxSum1=sum1;maxSum1Idx=i-3*k+1;}if(maxSum1+sum2>maxSum12){maxSum12=maxSum1+sum2;maxSum12Idx1=maxSum1Idx;maxSum12Idx2=i-2*k+1;}if(maxSum12+sum3>maxTotal){maxTotal=maxSum12+sum3;ans[0]=maxSum12Idx1;ans[1]=maxSum12Idx2;ans[2]=i-k+1;}sum1-=nums[i-3*k+1];sum2-=nums[i-2*k+1];sum3-=nums[i-k+1];}}return ans;}
}