KEY WORDS: [Rotated sorted array] [find target] [contain duplicates] publicbooleansearch(int[] nums,inttarget) {if(nums ==null|| nums.length == 0)returnfalse;intleft = 0;intright = nums.length - 1;while(left <=right) {intmid = left + (right - left) / 2;if(nums[mid] ==targ...
The method described in the theory performs binary search for arrays sorted in ascending order. Your task here is to modify the method such that: it allows searching in descending sorted arrays; it returns the first index of a target element from the beginning of the array (the leftmost index...
public TreeNode sortedArrayToBST(int[] nums) { return sortedArrayToBST(nums, 0, nums.length); } private TreeNode sortedArrayToBST(int[] nums, int start, int end) { if (start == end) { return null; } int mid = (start + end) >>> 1; TreeNode root = new TreeNode(nums[mid]...
Given an integer arraynumswhere the elements are sorted inascending order, convertit to aheight-balancedbinary search tree. Example 1: Input:nums = [-10,-3,0,5,9]Output:[0,-3,9,-10,null,5]Explanation:[0,-10,5,null,-3,null,9] is also accepted: Example 2: Input:nums = [1,3]...
Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 数组的中点肯定就是平衡BST的根节点,以此规律递归。 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; ...
Binary Search Tree Iterator 74 -- 1:31 App Leetcode-0081. Search in Rotated Sorted Array II 84 -- 2:53 App Leetcode-0144. Binary Tree Preorder Traversal 63 -- 4:35 App Leetcode-0101. Symmetric Tree 7 -- 1:29 App Leetcode-0167. Two Sum II - Input Array Is Sorted 12...
题目链接:https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/ 题目: Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 思路: 新建一个结点保存mid值,该结点的左右子树也递归生成,这是个常用的模板 ...
sorted order. With an array, inserting an item in the middle of the array requires that all remaining elements be shifted down one spot, and the array resized. In short, an array is usually a better choice if you have an idea on the upper bound of the amount of data that needs to ...
sorted order. With an array, inserting an item in the middle of the array requires that all remaining elements be shifted down one spot, and the array resized. In short, an array is usually a better choice if you have an idea on the upper bound of the amount of data that needs to ...
解析 高度平衡,保险起见,左右节点个数相同且均衡分配即可。二叉树问题还是递归问题,将数组分割成左右两个部分,然后递归求解即可。 伪代码 root=mid(nums)root.left=f(nums[:half])root.Right=f(nums[half+1:]) 代码 funcsortedArrayToBST(nums[]int)*TreeNode{iflen(nums)==0{returnnil}k:=len(nums)/2...