## 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...
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(...
int maxSubArray(std::vector<int>& nums) { int nums_size = nums.size(); return maxSubArray_div(nums,0,nums_size-1); } int maxSubArray_div(std::vector<int>& nums,int i,int j) { if(i>=j) return nums[i]; int h = (j+i)/2;//也可以使用移位操作 int left = maxSubArray_div...
(leetcode题解)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. 这是一道经典的题目,给定一个数...
解法一：暴力解法 思路：使用两层循环，时间复杂度是 O(n^2)。Python 代码：这种做法，虽然可以解决短的数列问题，但在提交时会超时。解法二：贪心+滑窗 思路：可以打败90%左右的用户，不过最经典的是分治法。最大子数列题，也是分治算法的经典应用。解法三：动态规划 第一步，定义临时数组，保存每个...
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 || 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....
Can you solve this real interview question? Maximum Product Subarray - Given an integer array nums, find a subarray that has the largest product, and return the product. The test cases are generated so that the answer will fit in a 32-bit integer. E
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...