Leetcode52.Maximum Subarray(贪心与分治)

题目描述:

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4], Output: 6 Explanation: [4,-1,2,1] has the largest sum = 6

题目分析:

该题属于Array类别,通过贪心可以实现O(n)级别的解决思路,通过分治法可以实现O(nlogn)的解决思路。

贪心法O(n)解决思路:

对于序列[A1,A2,...,AN]的遍历中,我们要求序列AI+...+AJ的和最大。

贪心算法在该问题的应用体现在我们在对序列遍历(i)求和的过程中，同时比较序列A0，...Ai的和与Ai的大小关系,两者的max值作为遍历至i时刻的当前序列最大sum.

并将当前序列最大sum与实际最大值比较，当序列遍历完毕，我们则求得整个序列的最大子集和。

Runtime: 72 ms, faster than 91.17% of Python3 online submissions for Maximum Subarray.

Memory Usage: 14.6 MB, less than 5.02% of Python3 online submissions for Maximum Subarray.

length = len(nums)summax = nums[0]cummax = nums[0]for num in nums[1:]:cummax = cummax+numif cummax <num:cummax = numif cummax >summax:summax = cummaxreturn summax

分治法O(nlogn)解决思路:

分治算法一般实现分为三个步骤:

(1)划分问题:将问题的实例划分为子问题,在本题中,就是将长度为l的序列不断划分为l/2的序列问题;

如划分中点m = l +int((r-l)/2)

在这里，需要注意区间的划分问题,在本题中,我采用较常用的左闭右开的区间划分方式，这样子在处理数组分割的问题比较自然。

对于一个长度为L的数组问题,我们视为区间[0,L)，在第一次递归分治的划分过程中,分成[0,m) [m,L)

区间划分问题决定了我们如何确定递归终止条件

如划分为左闭右开：

对于任何长度大于等于2的数组,我们的递归终止条件为left-right==1

对于长度为1的数组,我们直接设定递归终止条件为left==right

(2)递归求解:递归解决子问题,在本题中,通过递归调用maxsub(left,m) maxsub(right,m)，求得左右各子区间的最大序列sum值。递归终止条件在(1)中已提；

(3)合并问题:在(2)中我们已求得左右各子区间的最大序列sum值，接下来我们要考虑：左右子区间之间是否存在更大的序列sum值。

则我们需要从当前中点m为起点,向左遍历至left,求得最大的leftsummax

向右遍历至right-1，求得最大的rightsummax,

在将leftsummax+rightsummax与(2)中的最大值比较,作为当前递归的最后结果。

如果leftsummax+rightsummax大于(2)中的最大值，则说明左右子区间,包含中间m点在内的子序列存在更大的序列sum值

Runtime: 164 ms, faster than 5.83% of Python3 online submissions for Maximum Subarray.

Memory Usage: 14.8 MB, less than 5.02% of Python3 online submissions for Maximum Subarray.

class Solution:def maxsubhelper(self,nums,l,r):if r-l==1:return nums[l]if r==l:return nums[l]m =  l + int((r-l)/2)maxtmp = max(self.maxsubhelper(nums,l,m),self.maxsubhelper(nums,m,r))leftmax=0leftsummax = nums[m-1]for i in range(m-1,l-1,-1):leftmax += nums[i]leftsummax = max(leftsummax,leftmax)rightmax = 0rightsummax = nums[m]for i in range(m,r):rightmax += nums[i]rightsummax = max (rightsummax,rightmax)return max(maxtmp,leftsummax+rightsummax)def maxSubArray(self, nums: List[int]) -> int:return self.maxsubhelper(nums,0,len(nums))

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