vector<int> findDuplicates(vector<int>&nums) { vector<int> result;//use set if appear more than twiceif(nums.size() <2)returnresult;for(auto element : nums) {//element may be negative, don't use it directly as index//if (element < 0) element = -element;element =abs(element);if...
Runtime:35 ms, faster than100.00%of Java online submissions for Find Occurrences of an Element in an Array. Memory Usage:64.2 MB, less than100.00%of Java online submissions for Find Occurrences of an Element in an Array.
Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm’s runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. Example 1: Input: nums = [...
题目链接:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ 题目: Suppose a sorted array is rotated at some pivot unknown to you beforehand. 0 1 2 4 5 6 7might become4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 思...
Find the minimum element. You may assume no duplicate exists in the array. 思路:这个是找最小,和找指定数字其实差不多的。画个示意图吧 二分查找,关键是扔掉不包含最小值的那一半。 1.如果右半部分满足 n[m] < n[r] 则右半部分是递增的, 让 r = m 因为m位置有可能是最小的,把递增的那半扔掉...
leetcode-34. Find First and Last Position of Element in Sorte-binary-searchyjhycl 立即播放 打开App,流畅又高清100+个相关视频 更多 84 0 18:47 App leetcode-222-Counter Complete Binary Tree Nodes - binary search 2 0 10:00 App leetcode-852. Peak Index in a Mountain Array -binary-search...
这就是最基本的二分查找问题,对应于 LeetCode 35. 搜索插入位置:给定一个排序数组和一个目标值,在数组中找到目标值,并返回其索引。如果目标值不存在于数组中,返回它将会被按顺序插入的位置。我们已经分析过,target 按顺序插入的位置,就是满足 x≥ target 的第一个...
Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. ...
Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. 在一个升序的数组中,找出等于给定目标值首次出现后最后一次出现的位置,复杂度为O(log n),若果没有等于目标值的就返回[-1,-1] ...
Find the minimum element. You may assume no duplicate exists in the array. 数列基本有序,则使用二分查找。假设数列是n,s是起始下标,e是最后下标,m是中间元素下标。分为三种情况: ns < ne nm > ns > ne nm < ne < ns 情况1:ns < ne ...