In a given array nums of positive integers, find three non-overlapping subarrays with maximum sum.

Each subarray will be of size k, and we want to maximize the sum of all 3*k entries.

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:

Input: [1,2,1,2,6,7,5,1], 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.

Note:

• nums.length will be between 1 and 20000.
• nums[i] will be between 1 and 65535.
• k will be between 1 and floor(nums.length / 3).

首先用sliding window, 存下window下的sum.

然后类似于左右范围内的比较的问题，我们有一个left array和一个right array, 来存从左到右，left[i],到达i这个window的最大值。从右到左，right[i],到达i这个window的最大值
然后重新扫描这些window, (表示第三个的位置)，找到sum最大的地方。

class Solution {
    public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
        int sum = 0, len = nums.length - k + 1;
        int[] sums = new int[len];
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (i >= k) sum -= nums[i - k];
            if (i >= k - 1) sums[i - k + 1] = sum;
        }
        
        int[] left = new int[len];
        int best = 0;
        for (int i = 0; i < len; i++) {
            if (sums[i] > sums[best]) best = i;
            left[i] = best;
        }
        
        int[] right = new int[len];
        best = len - 1;
        for (int i = len - 1; i >= 0; i--) {
            if (sums[i] >= sums[best]) best = i;
            right[i] = best;
        }
        
        int[] res = {0, k, 2 * k};
        for (int j = k; j < len - k; j++) {
            int i = left[j - k], z = right[j + k];
            if (sums[i] + sums[j] + sums[z] > sums[res[0]] + sums[res[1]] + 
sums[res[2]]) {
                res[0] = i;
                res[1] = j;
                res[2] = z;
            }
        }
        return res;
    }
}