Python 代码: ## 用数组本身,来保存结果 class Solution(): ## 动态规划 def maxSubArray(self, nums): for i in range(1, len(nums)): if nums[i-1] > 0: ## 如果是正数,才会加到当前的位置 nums[i] += nums[i-1] return max(nums) ## 如果只有一个值,那么for循环不会启动,那么直接返回...
classSolution:defmaxSubArray(self, nums: List[int]) ->int: dp= [0foriin(range(len(nums)+1))] dp[0]= float("-inf")foriinrange(1,len(nums)+1): dp[i]= max(dp[i-1]+nums[i-1],nums[i-1])returnmax(dp)
python代码如下: 1classSolution:2#@param A, a list of integers3#@return an integer4defmaxSubArray(self, A):5#kadane's dynamic programming algorithm6maxSum = -1 * (1 << 30)7currSum =maxSum8forainA:9currSum = max(a, a +currSum)10maxSum =max(maxSum, currSum)11returnmaxSum...
RecursionError: maximum recursion depth exceeded while calling a Python object. 在讨论里面,我看到了精彩,这代码写得太漂亮了: class Solution: def maxSubArray(self, nums: List[int]) -> int: for i in range(1, len(nums)): nums[i] = max(nums[i-1]+nums[i], nums[i]) return max(nums...
class Solution(object): def maxSubArray(self, nums): """ :type nums: List[int] :rtype: int """ max_num=-2**31 this=0 for i in range(0,len(nums)): if this<0: this=0 this+=nums[i] max_num=max(max_num,this) return max_num 其实两句话可以概括,第一是存储,到当前的最大值...
[LeetCode&Python] Problem 53. Maximum Subarray Given an integer arraynums, 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],
本文将介绍计算机算法中的经典问题——最大子数组问题(maximum subarray problem)。所谓的最大子数组问题,指的是:给定一个数组A,寻找A的和最大的非空连续...
[Leetcode][python]Maximum Subarray/最大子序和 题目大意 由N 个整数元素组成的一维数组 (A[0], A[1],…,A[n-1], A[n]),这个数组有很多连续子数组,那么其中数组之和的最大值是什么呢? 子数组必须是连续的。 不需要返回子数组的具体位置。
the contiguous subarray [4,-1,2,1] has the largest sum = 6. click to show more practice. More practice: If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle. ...
Explanation: The result cannot be 2, because [-2,-1] is not a subarray. 问题本质: 本质:动态规划问题。 局部最优,全局最优。 product-乘法问题,存在的情况是 负数或者正数,或者从当前数开始新的连续元素相乘 可能发生的情况: 在某个节点,继续之前的增大/减小,从此节点转折。