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;
    }
}

结果：