LeetCode Longest Turbulent Subarray 131231
Given an integer arrayarr, returnthe length of a maximum size turbulent subarray ofarr. A subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray. More formally, a subarray[arr[i], arr[i + 1], ..., arr[j]]ofarris said to be turbulent...
[leetcode] 845. Longest Mountain in Array Description Let’s call any (contiguous) subarray B (of A) a mountain if the following properties hold: B.length >= 3 There exists some 0 < i < B.length - 1 such that B[0] < B[1] < … B[i-1] < B[i] > B[i+1] > … > B[...
解法3、双端队列 + 滑动窗口 classSolution {public:intlongestSubarray(vector<int>& nums,intlimit) {intn =nums.size(); deque<int> deq;//降序队列 记录最大值deque<int> inq;//升序队列 记录最小值intl =0, r =0, ans =0;while(r <n) {intx =nums[r];while(deq.size() && nums[deq.ba...
Given an array of integers nums and an integer limit, return the size of the longest continuous subarray such that the absolute difference between any two elements is less than or equal to limit. In case there is no subarray satisfying the given condition return 0. ...
链接:https://leetcode-cn.com/problems/longest-turbulent-subarray 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 思路是类似53题的动态规划。湍流子数组的定义是比较符号在相邻元素对之间翻转,那么我们可以定义两个数组 int[] increase 和 int[] decrease,都是以 nums[i] 为结尾的,最后...
Given an array of integersnumsand an integerlimit, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal tolimit. Example 1: Input: nums = [8,2,4,7], limit = 4 ...
方法一 方法二 Leetcode 978 问题描述 A subarray A[i], A[i+1], ..., A[j]ofAissaidtobe turbulentifandonlyif:Fori <= k < j, A[k] > A[k+1]whenkisodd,andA[k] < A[k+1]whenkiseven;OR,fori <= k < j, A[k] > A[k+1]whenkiseven,andA[k] < A[k+1]whenkisodd. ...
https://leetcode-cn.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/ 给你一个整数数组 nums ,和一个表示限制的整数 limit,请你返回最长连续子数组的长度,该子数组中的任意两个元素之间的绝对差必须小于或者等于 limit 。
解题思路:分别从左往右和从右往左边路nums,记录每个下标从左往右和从右往左的连续1的个数。最后计算出从左往右和从右往左的连续1的个数的和的最大值即可。 代码如下: classSolution(object):deflongestSubarray(self, nums):""":type nums: List[int] ...