## LeetCode 53 最大子数列和 Maximum Subarray class Solution(): def maxSubArray(self, nums): l = len(nums) dp = [0] * l ## 初始化数组全部为0 ## 套路第三步,初始特殊值为 nums 第一个元素 dp[0] = nums[0] #res_max = dp[0] ## 最终结果也初始化为 nums 第一个元素 for i in...
[LeetCode] Maximum Subarray Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array[−2,1,−3,4,−1,2,1,−5,4], the contiguous subarray[4,−1,2,1]has the largest sum =6. More practice: If you...
A subarray is a contiguous part of an array. 英文版地址 leetcode.com/problems/m 中文版描述 给你一个整数数组 nums ,请你找出一个具有最大和的连续子数组(子数组最少包含一个元素),返回其最大和。子数组 是数组中的一个连续部分。示例1:输入:nums = [-2,1,-3,4,-1,2,1,-5,4]输出:6解释:...
the contiguous subarray[4,−1,2,1]has the largest sum =6. 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. https://oj.leetcode.com/problems/maximum-subarray/ 思路1:Kadane算法,复杂度O(n...
leetcode || 53、Maximum Subarray problem: Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [−2,1,−3,4,−1,2,1,−5,4], the contiguous subarray [4,−1,2,1] has the largest sum = 6....
leetcode 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], Output:6 Explanation:[4,-1,2,1]has the largest sum =6....
LeetCode刷题日记 Day 28 Part 2 - Maximum Product Subarray, 视频播放量 70、弹幕量 0、点赞数 0、投硬币枚数 0、收藏人数 0、转发人数 0, 视频作者 blackwoodkane, 作者简介 ,相关视频:LeetCode刷题日记 Day 11 Part 2 - Unique Paths,LeetCode刷题日记 Day 24 Part 1
【CSON】LeetCode:718. Maximum Length of Repeated Subarray 0播放 · 总弹幕数02022-03-10 11:15:04 主人,未安装Flash插件,暂时无法观看视频,您可以… 下载Flash插件 点赞 投币收藏分享 稿件投诉 未经作者授权,禁止转载 官方网站:www.cspiration.com 微信号:cspiration01 微信公众号:北美CS求职 ...
the contiguous subarray [4,-1,2,1] has the largest sum = 6.中文:主要是给定一个数组,求解数组的子数组中,数组元素和最大的那一个子数组,返回的是最大子数组的和。2. 求解解法一最简单也是最容易想到的思路就是三层循环,对(i,j),i<=j的情况进行遍历,这种情况下的算法复杂度为O($n^3$)。代码如...
解决思路:将数组元素从左向右进行“滚雪球”式的相加,并和该位置原来的数组元素比较,选取最大值(核心思想也就是将所有元素都加一遍,选取所有相加结果里的最大值) /** * @param {number[]} nums * @return {number} */varmaxSubArray=function(nums){for(leti=1;i<nums.length;++i){nums[i]=Math.max...