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...
maxSubArray( nums[len(nums)//2: len(nums)] ) ## 计算跨终点数列的最大和: 拆成左边序列的最大和,加上右边序列的最大和 ## 从右到左,计算左子数组的总和 max_l = nums[len(nums)//2 - 1] ## 初始化为左区间最右边的那个值 cur_sum = 0 for i in range(len(nums)//2 - 1, -1, ...
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]<...
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-325-Maximum Size Subarray Sum Equals k,Error:Choosewrongalgorithm,nottwopointer.Weneedtousehashtabletodothis.Liketwosum,sincesum[i]-sum[j]=k=>sum[i]-k=sum[j].Sojustkeeptrackthesumandfindifsum[i]-...
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]<...
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...
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
(vector<int> &a, int l, int r) { if (l == r) { return (Status) {a[l], a[l], a[l], a[l]}; } int m = (l + r) >> 1; Status lSub = get(a, l, m); Status rSub = get(a, m + 1, r); return pushUp(lSub, rSub); } int maxSubArray(vector<int>& nums...
1186 Maximum Subarray Sum with One Deletion 删除一次得到子数组最大和 Description: Given an array of integers, return the maximum sum for a non-empty subarray (contiguous elements) with at most one element deletion. In other words, you want to choose a subarray and optionally delete one element...