heapsort

数据结构习题答案，线性链表的实现与适用，数据结构：思想与实现2-9(Data structure exercises answer, linear linked list implementations with applicable data structure: Ideas and Implementation 2-9)

最小生成树算法 克鲁斯法尔算法 球最佳路径(kulru method class maiun void cx )

数据结构中，运用链表实现四则括号运算等，运行在VC控制台上。(Data structure, the use of chain operations to achieve four brackets, etc., running in the VC console.)

给定n种物品和一背包。物品i的重量是wi>0,价值为vi>0,背包的容量为c。选择装入背包的物品，使得装入背包中物品的总价值最大。(Given n kinds of goods and a backpack. Weight of the item i is wi> 0, the value of vi> 0, knapsack capacity c. Select the items into the backpack, so into the backpack of items in the total value of the maximum.)

开括号和比括号的匹配，数据结构内容，很不错(More than brackets and brackets to open the match, the contents of data structure)

本代码主要实现了哈希表的设计，完成了哈希函数的相关要求。(The code is designed to achieve a hash table, the completion of the relevant requirements of the hash function.)

约瑟夫环（约瑟夫问题）是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。通常解决这类问题时我们把编号从0~n-1，最后结果+1即为原问题的解。 (Josephus (Josephus problem) is the application of a mathematical problem: Given n individuals (with numbers 1,2,3 ... n respectively) sitting around a round table. From the number of people gettin k, number of the m man out of the line he s the next person and a number of gettin number to m the man was out of the line and so the law is repeated until the round table were all out of the column. We numbered 0 ~ n-1, the final result is the original problem solution+1 usually solve these problems.)

实现约瑟夫问题的4种算法，并有计时功能进行效率测试(4 algorithms to resolve Josephus Problem with timing counter for testing efficiency)

背包算法的c++简单实现，很有用的，可以看看(Knapsack algorithm)

已知线性表中的元素以值递增有序排列，并以单链表作存储结构。试写一高效的算法，删除表中所有值大于mink且小于maxk的元素（若表中存在这样的元素）同时释放被删结点空间，并分析你的算法的时间复杂度（注意：mink和maxk是给定的两个参变量，它们的值可以和表中的元素相同，也可以不同）。(Known linear elements in the table to the value increments ordered, and a single list for the storage structure. Try to write a highly efficient algorithm to delete all values ​ ​ greater than mink and maxk elements smaller than (if there is such an element in the table) while the release of the deleted node space, and analyze your time complexity of the algorithm (Note: mink and maxk is given two parameters variables, their values ​ ​ and elements in the table can be the same or different).)