class Solution: def maximumSubarraySum(self, nums: List[int], k: int) -> int: # ans 维护所有长度为 k 且数字各不相同的子数组中,子数组和的最大值 ans: int = 0 # sum 维护当前滑动窗口 [l, r] 内的数字和 sum: int = 0 # num_to_cnt 表示滑动窗口 [l,
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...
you want to choose a subarray and optionally delete one element from it so that there is still at least one element left and the sum of the remaining elements is maximum possible.
Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn’t one, return 0 instead. Hashmap is also used in this problem. Iterate the array and calculate the sum, if map[sum-k] in the hashmap, update the maxlength=max(maxlength...
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]<...
Leetcode刷题 349.两个数组的交集 Intersection of Two Arrays 06:07 Leetcode刷题 219.存在重复元素 II Contains Duplicate II 04:57 Leetcode刷题 66.加一 Plus one 05:50 Leetcode刷题 53.最大子序和 Maximum Subarray 10:21 Leetcode刷题 62. 不同路径 Unique Paths 09:49 Leetcode刷题 198...
Explanation: The subarray[1, -1, 5, -2]sums to 3 and is the longest. 1. 1. 1. 首先计算出数组的前缀和,数组变为[1, 0, 5,3,6]。接着遍历这个前缀和的数组,用hashmap记录<prefix sum, i> -每个不同的前缀和和他们出现的位置。如果此时发现nums[i] - k在hashmap中,则res = Math.max(...
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...
(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...