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数据结构经典算法 约瑟夫问题 c++ 已经测试通过(josephus c++)

c语言完成的数据结构几个小算法游戏包括迷宫求解等(C language of complete data structure several small algorithm game includes maze solution, etc )

说明：  用链表的方式实现多项式的加减，其中结构体包含幂次和系数。(Polynomial addition and subtraction with a list of ways in which the structure and contains the power of coefficients.)

冒泡优化： 如果一个序列是int n[]={1,2,3,4,5,6,7,8,9} , 用正常的冒泡排序需要排8次才行，优化之后1次就好，也就是说序列越接近于正常序列，改进之后的冒泡排序的次数就越少，这样会给一个冒泡排序算法带了很大的效率。 思想：添加一个boolean变量用来判断冒泡是否是已经排好了顺序，如果boolean的值为false，说明是已经排好了，如果boolean的值true，说明没有排好，继续排。(If a sequence is int n [] = {, 1,2,3,4,5,6,7,8,9} need to row 8 times the job after optimization times like normal bubble sort, but alsomeans that the sequence is more close to the normal sequence, improved bubble sort, the less the number, which would give a bubble sort algorithm with a great deal of efficiency. Idea: add a boolean variable used to determine the bubble whether it is already lined up the order, if the boolean is false, indicating already lined up, if the boolean value of true, did not line up, continue to row.)

C++中，对链表的增删改查的基本使用，值得学习(C++, on the list of CRUD basic use, it is worth learning)

序列二次规划算法，能求解包含约束和无约束的问题，非常方便。(Sequential quadratic programming algorithm to solve constrained and unconstrained contain the problem, very convenient.)

数据挖掘中聚类算发中基于密度的算法DBSCAN(Hair count data mining clustering algorithm DBSCAN density-based)

建立链表的算法： 新节点链入表尾建立链表： 动态分配第一个节点变量； 用指针head 妥善保存链表第一个节点的地址值； 依次动态分配一个新节点变量； 将上一个节点的指针域指向新生成的节点； 循环3),4)步骤，直到所有节点均连入链表； 将最后一个节点的指针域赋值为0； (Establish the list of algorithms: The new node links footer build list: Dynamic allocation of the first node variable Keep the list head pointer of the first node address value In order to dynamically allocate a new node variable The domain of a node pointer to the new generation of nodes Cycle 3), 4) step, until all nodes are connected to the chain Pointer field will last node assigned to 0 )

kd-树（k维搜索树）是把二叉搜索树推广到多维数据的一种主存数据结构。这里给出了其代码的VC++实现(Classic data structures- KD tree full implementation, with a detailed explanation of the code, the novice data structure good materials)

在N*N的棋盘上(1<=N<=10)填入1,2,...N*N共N*N个数,使得任意两个相邻的数之和为素数. (In the N* N chessboard (1 < = N < = 10) filled 1,2, ... N* N total number of N* N so that any two adjacent numbers and prime numbers.)