Prim

迷宫求解一般采用“穷举法”，逐一沿顺时针方向查找相邻块（一共四块－东(右)、南(下)，西(左)、北(上)）是否可通，即该相邻块既是通道块，且不在当前路径上。用一个栈来记录已走过的路径栈是限定仅在表尾(top)进行插入或删除操作的线性表。(Maze solving generally use the " exhaustive" one by one to find adjacent blocks in a clockwise direction (a total of four- East (right), South (down), West (left), North (on)) whether to pass, that the both channel blocks adjacent blocks, and not in the current path. Use a stack to record the path traversed stack is limited only in the tail (top) to insert or delete a linear form.)

Huffman编码与译码，实现数据结构树的功能。(Huffman coding and decoding, data structure tree functions.)

说明：  标准答案习题1 1.1 简述C++语言程序的结构特点。 答： （1）C++程序由一个或多个函数组成，其中至少有一个主函数main()，程序从主函数开始执行，由主函数来调用其他函数。(Standard answer to Exercise 11.1 on C++ Language program structural characteristics. Answer: (1) C++ Procedures by one or more functions, of which at least one of the main function main (), the procedure started from the main function, from the main function to call the other function.)

排序程序,包含快速排序、归并排序、谢尔排序、插入排序。输出了比较次数和移动次数。(It is a sort program, including quick sort, merge sort, shell sort and insert sort.It outputs the compare times and movement times.)

数据结构一书上 关于迷宫游戏的算法 (Book on data structure algorithms on maze game)

吴伟民版数据结构每个算法的c实现方法，并模块化处理，将各个功能分开编程，有助于移植使用(failed to translate)

自己做的一个用用栈实现类的操作，是课内作业，希望和大家互相交流(Own a type used by the realization of the operation of the stack is class work, hope and we can exchange ideas)

2008软件公司面试题:排序大全：简单选择排序、冒泡排序、选择排序、shell排序、快速排序、插入排序、堆排序(从小到大)、归并排序（利用递归）(2008 software companies face questions: Sort Guinness: simple choice to sort, bubble sort, select sort, shell sort, quick sort, insert sort HEAPSORT (from small to large), Merging Sort (recursive))

求二叉树的深度: 给定一个有根二叉树，规定连接两个节点的每条边长度是1，定义树的深度为根到叶子节点距离的最大值。 (Description 给定一个有根二叉树，规定连接两个节点的每条边长度是1，定义树的深度为根到叶子节点距离的最大值。 )

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 )