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(...
the contiguous subarray[4,−1,2,1]has the largest sum =6. 思路:初始化sum和max值为数组首元素值,从头开始遍历数组,sum<0则将sum置为0,sum>max则更新max为sum。 1classSolution {2public:3intmaxSubArray(intA[],intn) {4intsum=A[0],max=sum;5for(inti=1;i<n;i++)6{7if(sum<0)8sum=...
思路2:Prefix Sum 扫描一遍数组的过程中记下当前的前缀和,以及最小的前缀和,两者相减就是当前能够获得的最大子数组和。 classSolution{public:intmaxSubArray(vector<int>& nums){intn = nums.size();if(n ==0)return0;intret = nums[0], sum =0, min_sum =0;for(inti =0; i < n; ++i) { s...
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...
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,-...
我们将每种subarray的可能性都算进去,并一直更新出现过的最大值。很明显这种解法实在是太慢了!!!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 =...
1publicclassSolution2{3publicintmaxSubArray(int[] nums)4{5//Solution 1: O(n)6//check param validation.7if(nums ==null|| nums.length == 0)8return0;910intsum = 0;11intmax =Integer.MIN_VALUE;1213//iterate nums array.14for(inti = 0; i < nums.length; i++)15{16//choose a larg...
## 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...
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 ...
publicclassSolution {publicintMaxSubArray(int[] nums) {if(nums.Length ==0)return0;intmax =int.MinValue;for(inti =0; i < nums.Length; i++){inttempSum=0;for(intj =i; j < nums.Length; j++){ tempSum+=nums[j];if(tempSum >max) ...