[leetcode] 53. Maximum Subarray @ python

最新推荐文章于 2022-01-31 01:28:45 发布

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最新推荐文章于 2022-01-31 01:28:45 发布

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本文链接：https://blog.youkuaiyun.com/danspace1/article/details/86179406

本文深入探讨了LeetCode上经典题目“最大子数组和”的两种高效解法：Kadane’s算法和动态规划。通过实例讲解，展示了如何在O(n)时间复杂度内找到连续子数组的最大和，适合算法初学者和进阶者学习。

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原题

https://leetcode.com/problems/maximum-subarray/
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

使用Kadane’s 算法, 遍历数组, 计算在i点时的最大和 max_end_here, 并更新max_so_far, https://en.wikipedia.org/wiki/Maximum_subarray_problem#Kadane’s_algorithm
Time: O(n)
Space: O(1)

代码

class Solution:
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        max_end_here = max_so_far = nums[0]
        for i in range(1, len(nums)):
            max_end_here = max(nums[i], nums[i] + max_end_here)
            max_so_far = max(max_so_far, max_end_here)
        return max_so_far

解法2

动态规划, 初始化dp数组, dp[i]表示在i点最大的子数组. 状态转移方程:

dp[i] = dp[i-1] + nums[i] if dp[i-1] > 0 else nums[i]

然后我们求dp中最大的值即可.

代码

class Solution:
    def maxSubArray(self, nums: 'List[int]') -> 'int':
        dp = nums[:]
        for i in range(1, len(nums)):
            if dp[i-1] > 0:
                dp[i] = dp[i-1] + nums[i]
            else:
                dp[i] = nums[i]
                
        return max(dp)