二分法基本上学计算机的都听过,但是有人不知道的就是其实二分法是减治法的思想. 所谓减治法和分治法有一个主要差别就是减治法是减去一般,就是分治之后只需要解决原问题的一半就可以了得到全局问题的解了.所以速度很快. 下面是二分法的递归程序和非递归程序和主测试程序: #include<iostream> #include<vector> using namespace std; template<typename T> int recurBiSearch(const vecto…

Binary Search 二分法方法总结 code教你做人:二分法核心思想是把一个大的问题拆成若干个小问题,最重要的是去掉一半或者选择一半. 二分法模板: public int BinarySearchTemplate(int[] nums,int target) { if(nums == null || nums.length == 0) return -1; int lo = 0; int hi = nums.length - 1; //A: lo < hi [1,2]找1 找last p…

［抄题］: Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target. [暴力解法]: 时间分析: 空间分析: ［奇葩输出条件］: ［奇葩corner case］: ［思维问题］: 以为要用主函数+ DFS来做.错了,“最近”还是直接用二分法左右查找得了 ［一句话思路］: 定义一个res,如果root离target的距离小 就替换…

Leetcode之二分法专题-704. 二分查找(Binary Search) 给定一个 n 个元素有序的(升序)整型数组 nums 和一个目标值 target  ,写一个函数搜索 nums 中的 target,如果目标值存在返回下标,否则返回 -1. 示例 1: 输入: nums = [-1,0,3,5,9,12], target = 9 输出: 4 解释: 9 出现在 nums 中并且下标为 4 示例 2: 输入: nums = [-1,0,3,5,9,12], target = 2 输出:…

与排序算法不同,搜索算法是比较统一的,常用的搜索除hash外仅有两种,包括不需要排序的线性搜索和需要排序的binary search. 首先介绍一下binary search,其原理很直接,不断地选取有序数组的组中值,比较组中值与目标的大小,继续搜索目标所在的一半,直到找到目标,递归算法可以很直观的表现这个描述: int binarySearchRecursive(int A[], int low, int high, int key) { ; ; , key); , high, key); e…

Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST. No…

Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST. No…

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL. For exampl…

Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST. Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than or equal to the nod…

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL. For exampl…