资源列表
c
- 顺序表的创建、查找和删除_silas_新浪博客,silas,顺序表,元素,查找,线性表,程序代码,it-Create a sequence table, find and delete _silas_ Sina blog, silas, order form, the elements, find, linear tables, program code, it
shuzu1
- 一位数组的源代码 显示顺序表 输出顺序表的长度 判断顺序表是否为空表插入数据 删除数据\查找数 显示菜单 清空 退出-An array of source code shows whether the length of the table to determine the order of the sequence table output sequence table is an empty table to insert data delete data \ find the number
lianbiao
- 链表源代码 重新录入表 插入到第一个结点 查找指定结点直接前驱-Table lists the source code re-entry into the first node to find the specified node direct precursor
PrjEGov42
- 非常好用的排课系统源码 可以参考一下 然后试着-Timetable system very easy to use source code can refer to and then try to learn
bank-data
- 数据挖掘常用的银行数据，很适合初学者，和大家一起分享！-Data mining commonly used bank data, it is suitable for beginners, and share with everyone!
2
- 找出给定序列的所有逆序对，并输出。 逆序对：若i < j ==> A[i] > A[j] 则(i,j)为一个逆序对-Identify all the given sequence in reverse order right, and output. On the reverse: if i < j ==> A [i]> A [j] then (i, j) as a reverse pair
3
- 利用random(0,1)实现random(a,b) 原型: void srand(unsigned seed) 功能: 产生随机数的起始发生数据，和rand函数配合使用-Starting with the use of a random number generated occurrence data, and the rand function: the use of random (0,1) to achieve random (a, b) prototype: void sra
4
- 给定k个链表，利用最小堆实现有序链表的k路合并，并输出结果。-Given k a list using the minimum heap achieve an orderly list of k-way merge, and outputs the result.
7
- 实现：平均排序 (k排序) k排序的定义：A[i] <= A[i + k] 即为希尔排序的不完全实现-Implementation: Average sort (k sorting) k sort defined: A [i] < = A [i+ k] is not fully realized Hill sort
least-common-ancestor
- Tarjan的脱机最小公共祖先问题，里面使用了pair,map以及结构体等相关结构-Tarjan off-line least common ancestors problem
rbTree
- 红黑树的插入删除，旋转，前驱和后继等操作都有实现。-in this source file ,we have realized the insertion,deletion and rotation,predecess and successor operations
introduction-to-algorithm
- 算法导论第三版书里的部分代码（大约到第22章，22章之后的代码稍后会更新）。-introduce to algorithm,this compressed file includes all the source code before chapter 22 (including 22), which all are writen by myself.