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 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(...
## 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: def maxSubArray(self, nums: List[int]) -> int: if sum(nums[0:])<=sum(nums[:-1]): return sum( self.maxSubArray(nums[:-1]) ) elif sum(nums[1:])>=sum(nums): return sum(self.maxSubArray(nums[1:])) else: return sum(self.maxSubArray(nums[1:-1])) ...
*/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...
class Solution{public:intmaximumSum(vector<int>&arr){intn=arr.size(),result=arr.front();vector<int>zero(n),one(n);zero.front()=one.front()=arr.front();for(inti=1;i<n;i++){zero[i]=max(arr[i],zero[i-1]+arr[i]);one[i]=max(zero[i-1],one[i-1]+arr[i]);result=max(...
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. 【题目分析】 题目要求我们在给定的一个数组中找出一个数组元素之和最大子数组,并返...
class Solution { public: int maxSubArray(vector<int>& nums) { //return maxSubArray_dp_normal(nums); return maxSubArray_dp_seleted(nums); } int maxSubArray_dp_normal(vector<int>& nums) { const int nums_size = nums.size(); vector<int> dp = nums;//dp[i]表示以nums[i]结尾的和最大...
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. 翻译: 给你一个整数型数组nums,找出连续的子数组(长度至少为1)使得其和最大,返回这个最大值。
classSolution {public:intmaxSubArray(vector<int>&nums) {if(nums.size()==0)return0;returnmaxSubArray(nums,0, nums.size() -1); }//l代表数组左端low,h代表数组右端high,返回最大sumintmaxSubArray(vector<int>& arr,intl,inth) {//Base Case: Only one elementif(l ==h)returnarr[l];//Find...
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