题目如下：
In a given array
numsof 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 all3*kentries.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.lengthwill be between 1 and 20000.nums[i]will be between 1 and 65535.kwill be between 1 and floor(nums.length / 3).
解题思路：本题如果只要求求出三段子数组的和的最大值，那会简单很多。记total[i]为arr[i:i+k]段的和，dp_left_max[i]为nums[:i]区间内长度为k的子数组的和的最大值，dp_right_max[i]为nums[i:len(nums)]区间内长度为k的子数组的和的最大值，很显然如果中间段的子数组的下标为k，那么可以得到三段和的最大长度的表达：total[i] + dp_left_max[i-k] + dp_right_max[i+k] 。只要遍历数组arr，即可求出最大值。求出后就是计算出左边以及右边最大值出现时的最小下标，这个可以通过二分查找实现。
代码如下：
class Solution(object): def maxSumOfThreeSubarrays(self, nums, k): """ :type nums: List[int] :type k: int :rtype: List[int] """ count = sum(nums[:k]) total = [count] total_inx = {} total_inx[count] = [0] dp_left_max = [count] dp_left_max_count = count for i in range(k, len(nums)): count -= nums[i - k] count += nums[i] total += [count] total_inx[count] = total_inx.setdefault(count,[]) + [i-k + 1] dp_left_max_count = max(dp_left_max_count,count) dp_left_max.append(dp_left_max_count) reverse_num = nums[::-1] count = sum(reverse_num[:k]) dp_right_max = [count] dp_right_max_count = count for i in range(k, len(reverse_num)): count -= reverse_num[i - k] count += reverse_num[i] dp_right_max_count = max(dp_right_max_count,count) dp_right_max.insert(0,dp_right_max_count) #print total #print total_inx #print dp_left_max #print dp_right_max max_sum = -float(‘inf‘) mid_inx = 0 left_val = 0 right_val = 0 for i in range(k,len(nums)-k-k+1): count = total[i] + dp_left_max[i-k] + dp_right_max[i+k] if count > max_sum: mid_inx = i left_val = dp_left_max[i-k] right_val = dp_right_max[i+k] max_sum = count #print left_val,mid_inx,right_val left_inx = total_inx[left_val][0] import bisect right_inx = bisect.bisect_left(total_inx[right_val],mid_inx+k) return [left_inx,mid_inx,total_inx[right_val][right_inx]]