Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.A subarray is a contiguous part of an array. 英文版地址 leetcode.com/problems/m
leetcode根本不让你通过的,比如100个数据100^100次循环 天知道要跑多久。 神奇的Kadane's algorithm 代码 class Solution: def maxSubArray(self, nums: List[int]) -> int: max_current = max_global = nums[0] for num in nums[1:]: max_current = max(num,max_current+num) if(max_current>...
*/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...
class Solution { public: int maxSubArray(int A[], int n) { if (A == NULL || n < 1) { return INT_MIN; } int contsum = A[0]; int maxsum = A[0]; for (int i=1; i<n; i++) { contsum = max(contsum + A[i], A[i]); maxsum = max(maxsum, contsum); } return...
classSolution {public:intmaxSubArray(vector<int>&nums) {intlen=nums.size();intmax_ending_here=nums[0];intmax_so_far=nums[0];for(inti=1; i<len; i++) { max_ending_here=max(nums[i],max_ending_here+nums[i]); max_so_far=max(max_so_far,max_ending_here); ...
https://leetcode.com/problems/maximum subarray/ 暴力方法:枚举所有子数组,计算子数组和,记下最大的。复杂度是O(n^3)。 思路1:Dynamic Programming 定义f[i]表示以nums[i]结尾的子数组的最大长度。则有 由于记忆深度只有1
public int maxSubArray(int[] A) { return maxSumDAC(A, 0, A.length-1); } } 两种遍历方法: public class Solution { public int maxSubArray(int[] A) { int maxSum = Integer.MIN_VALUE; for(int i=0; i<A.length; i++) //通过三重循环,穷举所有可能子数组的和 ...
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,-...
public class Solution { public int maxSubArray(int[] nums) { int max = Integer.MIN_VALUE;//设置最小值 int sum = 0;//每一个分组的和 int i = 0; while(i < nums.length){ sum += nums[i];//每一个分组的前n项和 if(max < sum){ ...