class MaximumSubarrayPrefixSum { public int maxSubArray(int[] nums) { int len = nums.length; int maxSum = Integer.MIN_VALUE; int sum = 0; for (int i = 0; i < len; i++) { sum = 0; for (int j = i; j < len; j++) { sum += nums[j]; maxSu...
*/publicclassSolution53{publicstaticvoidmain(String[] args){Solution53solution53=newSolution53();int[] arr = {-2,1, -3,4, -1,2,1, -5,4}; System.out.println(solution53.maxSubArray(arr)); }/** * maxSum 必然是以nums[i](取值范围为nums[0] ~ nums[n-1])结尾的某段构成的,也就...
## 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...
dp[i] represents the maximum sum of subarray which ends in nums[i], and dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]). and since we have to include nums[i] due to it’s on the defination of dp[i], and when dp[i-1]<0 we can just choose not to add it. so the...
如果每次得到新的candidate都和全局的maxSum进行比较,那么必然能找到最大的max sum subarray. 在循环过程中,用maxSum记录历史最大的值。从nums[0]到nums[n-1]一步一步地进行。 思路二: 遍历array,对于每一个数字,我们判断,(之前的sum + 这个数字) 和 (这个数字) 比大小,如果(这个数字)自己就比 (之前的su...
LeetCode 53. Maximum Subarray 程序员木子 香港浸会大学 数据分析与人工智能硕士在读 来自专栏 · LeetCode Description 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,-...
Leetcode 53. Maximum Subarray 2. Solution **解析:**Version 1,简单粗暴,前i个元素总和大于0,则这些元素对总和是有贡献的,要保留,否则,则丢弃前i个元素。重新开始执行上述操作,每次加完记得更新最大值。Version 2,采用动态规划求解,首先定义状态,dp[i]是以nums[i]为结尾的连续子数组的最大和,状态转移方程...
**Leetcode 53 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....
Leetcode No.53 Maximum Subarray(c++实现) 1. 题目 1.1 英文题目 Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. 1.2 中文题目 给定一个整数数组 nums ,找到一个具有最大和的连续子数组(子数组最少包含一个元...
LeetCode53 Maximum sum of subarray classic dp: dp[i] represents the maximum sum of subarray which ends in nums[i], and dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]). and since we have to include nums[i] due to it’s on the defination of dp[i], and when dp[i-1]<...