【leetcode】689. Maximum Sum of 3 Non-Overlapping Subarrays

题目如下：

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).

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]]

原文地址：https://www.cnblogs.com/seyjs/p/11616699.html