三个无重叠子数组的最大和。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。

时间紧，见代码。

代码用golang编写。代码如下：

package main

import "fmt"

func main() {
    nums := []int{1, 2, 1, 2, 6, 7, 5, 1}
    k := 2
    ret := maxSumOfThreeSubarrays(nums, k)
    fmt.Println(ret)
}

func maxSumOfThreeSubarrays(nums []int, k int) []int {
    N := len(nums)
    range2 := make([]int, N)
    left := make([]int, N)
    sum := 0
    for i := 0; i < k; i++ {
        sum += nums[i]
    }
    range2[0] = sum
    left[k-1] = 0
    max := sum
    for i := k; i < N; i++ {
        sum = sum - nums[i-k] + nums[i]
        range2[i-k+1] = sum
        left[i] = left[i-1]
        if sum > max {
            max = sum
            left[i] = i - k + 1
        }
    }
    sum = 0
    for i := N - 1; i >= N-k; i-- {
        sum += nums[i]
    }
    max = sum
    right := make([]int, N)
    right[N-k] = N - k
    for i := N - k - 1; i >= 0; i-- {
        sum = sum - nums[i+k] + nums[i]
        right[i] = right[i+1]
        if sum >= max {
            max = sum
            right[i] = i
        }
    }
    a := 0
    b := 0
    c := 0
    max = 0
    for i := k; i < N-2*k+1; i++ { // 中间一块的起始点 (0...k-1)选不了 i == N-1
        part1 := range2[left[i-1]]
        part2 := range2[i]
        part3 := range2[right[i+k]]
        if part1+part2+part3 > max {
            max = part1 + part2 + part3
            a = left[i-1]
            b = i
            c = right[i+k]
        }
    }
    return []int{a, b, c}
}

执行结果如下：