【LeetCode】53. Maximum Subarray 解题报告（Python）

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本文链接： https://blog.csdn.net/ttinch/article/details/102566778

题目地址：https://leetcode.com/problems/maximum-subarray/

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.

Follow up:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

解法1：动态规划

$\begin{matrix} dp[i] = dp[i-1] + s[i] & dp[i-1] \geq 0\\ dp[i] = s[i] & dp[i-1] < 0 \end{matrix}.$

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        p = [0 for i in range(len(nums))]
        p[0] = nums[0]
        max_num = p[0]
        for i in range(1,len(p)):
            p[i] = p[i-1] * (p[i-1] > 0) + nums[i]
            max_num = max(max_num, p[i])
        return max_num

解法2：分治法

分治的策略：
将数组均分为两个部分，那么最大子数组会存在于：

• 左侧数组的最大子数组
• 右侧数组的最大子数组
• 左侧数组的以右侧边界为边界的最大子数组+右侧数组的以左侧边界为边界的最大子数组

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        return self.solve(nums, 0, len(nums)-1)

    def solve(self, nums: List[int], low: int, high: int) -> int:
        if low == high:
            return nums[low]
        mid = low + int((high-low)/2)
        leftMax = self.solve(nums, low, mid)
        rightMax = self.solve(nums, mid+1, high)
        leftSum = nums[mid]
        tmp = nums[mid]
        for i in range(mid-1, low-1, -1):
            tmp += nums[i]
            leftSum = max(leftSum, tmp)
        rightSum = nums[mid+1]
        tmp = nums[mid+1]
        for i in range(mid+2, high+1):
            tmp += nums[i]
            rightSum = max(rightSum, tmp)
        return max([leftMax, rightMax, leftSum+rightSum])

分治法和动态规划区别

• 子问题是否重叠

分治法是指将问题划分成一些独立地子问题，递归地求解各子问题，然后合并子问题的解而得到原问题的解。与此不同，动态规划适用于子问题独立且重叠的情况，也就是各子问题包含公共的子子问题。