lc 689. Maximum Sum of 3 Non-Overlapping Subarrays

https://leetcode.com/problems/maximum-sum-of-3-non-overlapping-subarrays/description/

 

给一个array，一个k

找到三段连续k长的子数组，使和最大。

为什么是3，不是4，5。三可以往左往右看，找最大。就是本身算一个，左边最大，右边最大。就知道固定当前情况下的最大值。这样遍历一遍就行了。

code：

class Solution:
    def maxSumOfThreeSubarrays(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        s=sum(nums[:k])
        sms=[s]
        for j in range(k,len(nums)):
            i=j-k
            s+=nums[j]-nums[i]
            sms.append(s)
        smsl=[(sms[0],0)]
        for i in range(1,len(sms)):
            smsl.append(smsl[-1] if smsl[-1][0]>=sms[i] else (sms[i],i))
        smsr=[[-1,0] for i in range(len(sms))]
        smsr[-1]=[sms[-1],len(sms)-1]
        for i in reversed(range(len(sms)-1)):
            if smsr[i+1][0]>sms[i]:
                smsr[i][0]=smsr[i+1][0]
                smsr[i][1]=smsr[i+1][1]
            else:
                smsr[i][0]=sms[i]
                smsr[i][1]=i
        maxx=-1
        p1=-1
        p2=-1
        p3=-1
        for i in range(k,len(smsl)-k):
            if sms[i]+smsl[i-k][0]+smsr[i+k][0]>maxx:
                p1=smsl[i-k][1]
                p2=i
                p3=smsr[i+k][1]
                maxx=sms[i]+smsl[i-k][0]+smsr[i+k][0]

        return[p1,p2,p3]