[LeetCode] 34. Find First and Last Position of Element in Sorted Array == [LintCode] 61. Search for a Range_Easy tag: Binary Searchfirst index and last index == target in duplicates. [LeetCode] 35. Search Insert Position_Easy tag: Binary Searchfirst index >= target, without duplicates...
Similar Questions Search in a Sorted Array of Unknown Size Medium Maximum Count of Positive Integer and Negative Integer Easy Discussion (152) Copyright ©️ 2025 LeetCode All rights reserved 12.5K 152 0 Online nums = [-1,0,3,5,9,12] ...
关键点:二分查找 classSolution {publicintsearch(int[] nums,inttarget) {intleft = 0;intright = nums.length - 1;while(left <=right) {intmid = left + (right - left) / 2;if(nums[mid] ==target)returnmid;elseif(nums[mid] < target)//if mid is less than target, move to right half...
Input:root = [4,2,7,1,3], val = 5Output:[] Constraints: The number of nodes in the tree is in the range[1, 5000]. 1 <= Node.val <= 107 rootis a binary search tree. 1 <= val <= 107 FindTabBarSize FindBorderBarSize
LeetCode Top 100 Liked Questions 96. Unique Binary Search Trees (Java版; Medium) 题目描述 Given n, how many structurally unique BST's (binary search trees) that store values 1 ...n? Example: Input: 3 Output: 5 Explanation: Given n = 3, there are a total of 5 unique BST's: ...
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LeetCode Top 100 Liked Questions 236. Lowest Common Ancestor of a Binary Tree (Java版; Medium) 题目描述 Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined betw...
Demonstrate all the questions on LeetCode in the form of animation.(用动画的形式呈现解LeetCode题目的思路) - binary2010/LeetCodeAnimation
来自专栏 · LeetCode刷题记录 98. Validate Binary Search Tree 难度:medium Given a binary tree, determine if it is a valid binary search tree (BST).Assume a BST is defined as follows:The left subtree of a node contains only nodes with keys less than the node's key.The right subtree ...
3 ii.If the second max element in the current search space is located in the first half of the array (i.e.mid > smax_all), then find the second maximum element in the first half (i.e.smax_1st = object.query(left, mid)). If the second max elements in the ent...