1. Problem Descriptions:Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.If target is not found in the array, return [-1…
LeetCode刷题系列—34. Find First and Last Position of Element in Sorted Array 1.题目描述 英文版: 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 ...
- LeetCodeleetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/ 解题思路 1. 二分找元素,双指针找区间 class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { vector<int> res(2, -1); int pos = bsearch(nums, target, 0, nums.size...
Find First and Last Position of Element in Sorted Array 解题思路 数组查找问题,考虑采用二分查找降低时间复杂度,找到target后只需分别向前向后找到start和end即可。 代码 class Solution: def searchRange(self, nums: List[int], target: int) -> List[int]: low, start = 0, -1 high, end = len(n...
LeetCode-Find First and Last Position of Element in Sorted Array,LeetCodeJavaC++FindFirstandLastPositionofElementinSortedArray
链接:https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array 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。 这道题的最优解是二分法。思路是通过二分法分别找到第一个插入的位置和第二个插入的位置。注意找第一个位置和第二个位置的不同,两者都是...
给定一升序整数数组,找出给定整数的起止边界。算法时间复杂度要为O(log n) 思路:二叉搜索 Runtime: 4 ms, faster than 65.40% of Java online submissions for Find First and Last Position of Element in Sorted Array. Memory Usage: 30.5 MB, less than 28.29% of Java online submissions for Find First...
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]. ...
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...
这道题应该是属于比较明显的用二分法的题。首先通过二分法找到该元素在数组中首次出现的位置,若通过一次查找该元素没有出现在该列表中,则直接返回[-1, -1]。找到了first position后,从该位置到列表末尾这个区间再次通过二分法寻找该元素最右边的位置。代码如下,主要参考: ...