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原文：http://www.cppblog.com/guogangj/archive/2009/10/26/99502.html十一、平衡二叉查找树（AVL）前面说了，二叉查找树方便查找取值插入删除，其复杂度不过为Ο(logn)，但这是个“期望值”，因为我们也有比较差的情况，比如

原文：http://hi.baidu.com/jiang_yy_jiang/blog/item/931d763ed8edc8f2838b13c3.html 已知一组关键字为(26，36，41，38，44，15，68，12，06，51)，用除余法构造散列函数，用线性探查法解决冲突构

原文：http://blog.163.com/clevertanglei900@126/blog/static/111352259201131891452434/1 快排效率是不稳定的nlogn2 二叉树实现排序的效率是稳定的nlogn3 用二叉树实现排序有两种方法： 二叉排序树

1.确定区间[a,b] , 验证f(a) * f(b) 2.求区间(a,b)的中点m3.计算f(m)3.1 若f(m) = 0 , m 就是函数的零点3.2 if f(a) * f(m) 3.3 else if f(b) * f(m) 4.if abs(a - b)  重复2-4

In Balance Tree the point is how to balance the tree byrotating .Two Things to be find out :1>    ClosetLost Balance Point -- LBP ( Abs(leftHeight – rightHeight) > 1 )2>    Find Sub-Tree --S

Red-Black Tree  5 essentials :·        each node color is red or black .·        root color is black·        each leaf is black·        if parent is red color , then children are black .

huffMan Tree:Imagine that here is one article , which contains 27 'A'  , 8 'B' ...and 5 'F' :A 27 , B 8 , C15, D15, E30 ,F5So now build a Huffman tree(1)take first TWO numbers , (2)constru

For each some Graph , may need to find out the "Minimum cost " way to build .for this graph , Now need to find out the min-cost tree .This method is named asKruskal :Now we get {v0,v1,

How To Get Min-Cost Between twopoints in graph(Dijkstra’s algorithm) Now See This Graph :which is just  like asubway map . Now Question is : how to calculate  fromA to (B,C,D,E,F

CPM(Critical Path Method ) is now widely used in projectmanagement to control the schedule of the project life cycle . Now I will show you how to find out the critical activities andhow to create

归并排序本文章将介绍归并排序，时间复杂度为n*lg(n)的归并排序，不仅效率高，而且是稳定的（而堆排序，快排是不稳定的）。要排序的数组：5 , 1, 9 ,3, 7 , 4 , 8 ,6 , 2 归并排序过程：1.      计算组元素个数,count ,以count=2为起始个数，每次count*=2,count∈[2,arr.Count]2.      每次把数

二分查找，插值查找，斐波那契查找：1.二分查找：伪代码：while(leftbeginif key == arr[left] or key == arr[right] or key == arr[left+right/2]return indexelse if key< arr[(left+index)/2]then right = (left+right)/2

ToopoLOgical Sort Target :Sort the nodes in graph ,to find out the graph have cycle or not .result nodes only contains the nodes have not "involed in the cycle"steps :0.prepare node

快速排序 快速排序使用分治的思想，以某个数为分割点，每一趟排序下来，左边数都比它小，右边数都比它大，然后再对左右两边数组做递归操作。时间复杂度(平均情况)：n*lg(n) 排序步骤：1.      以左边首元素为分割点，拿出来存在midElement里（这时候左边首位置空了）2.      从右边找比midElement小的数，放在刚才左边空的位置，就是首位（这时，

堆排序本文将介绍一下堆排序的过程。 要点：（最小堆为例）1.      调整堆函数：传入节点索引，对比当前节点与父节点大小，如果小于则交换位置，以父节点位置递归执行。2.      设index=Arr.Count/ 2 -1 ,取到index位置的子节点，将子节点和当前节点依次传入递归函数执行，i∈[0,index],i--循环执行3.      每次移除根节点，

排序详解 之 希尔排序

排序详解 之 选择排序

Set Deepth for each tree node of a forestTreeNode Entity: public class TreeNode    {        public string Name { get; set; }        public TreeNode ParrentNode { get; set; }

转自：http://www.cnblogs.com/ycsfwhh/archive/2010/12/20/1911232.htmlprivate const double EARTH_RADIUS = 6378.137;//地球半径private static double rad(double d){   return d * Math.PI / 180.0;

1.声明队列Expressions，和堆栈Operators ，并定义优先级：( 和 ):3* 和/ : 2+ 和 - :1 数值越大，优先级越高foreach(每一个计算单元 in 当前中缀表达式){2.遇到数字直接进队列3.遇到左括号 '('直接进堆栈4.遇到右括号while(栈顶元素!=左括号){入队列弹出}弹出（为

2-3Tree  Defination :1.    A tree can have more than one values2.    If node have 1 value ,then only 2 children3.    If node have 2 value ,then have 3 children During Insertion ,

/////////////////Node.h//////////////////class BTree{public:    ////methods    void InsertNode(TNode* node);    void PrintNode(TNode* head);

using System;using System.Collections.Generic;using System.Linq;using System.Text;namespace Imps.Services.IDCService.Utility{

public List GetRandomLst(int from, int to)        {            var lst = Enumerable.Range(from, to - from + 1);            return lst.OrderBy(a => Guid.NewGuid()).ToList();        }

/////定义////////////////////////////节点类class Node{public:    ////methods    Node(void);    Node(int data);    ~Node(void);    /////members

复习二分查找法。public int FindPosition(int num, int[] arr)        {            int left = 0;            int right = arr.Length - 1;

////欧几里得算法(求两个数的最大公约数)int God(int M,int N){int rem ;while(N > 0){rem=M%N;M=N;N=rem;}return M;} 调用： int _tmain(int argc, _TCHAR* argv[]){    int rlt = God(360,60);

求最大子序列的和 传统的嵌套循环的解法：  /////获得最大子序列的和,复杂度O(n^2)int GetMaxSubSequence(int* arr,int len){    int i, j;    int curSum; /* 当前序列和 */    int maxSum; /* 最大序列和 */    /* 初始化最大子序列和

问题描述：Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alph

Non--Abundant numbers

硬币面值的组成多少种可能---Javascript实现

Project Ruler 算法练习之除数问题

Project Ruler 算法练习之 10 进制 转 2进制 以及数字对称

Project Ruler 算法练习之 Truncate Prime

两大数相乘 -- javascript 实现

var indexOfSorted = function f(arr,n){ //assume : arr has been sorted var low = 0; var high = arr.length - 1; var mid = (low + high) / 2; while(high - low > 1){ if(n == arr[low]){return low

算法之会议时间安排

Step 1: n ==1 : return 1 n == 2 : return  [1,1],[2]Step 2:for n > 2a.arr.push(n)b.arr.push([n-1,1])c.1 get result of recursion(n-2)c.2 combine n==2 & result => retc.3 remove duplicate record in retcod

Problem :It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and

var arr = new Array();for(var i = 0 ;i < 6 ; i++){if(i == 0){arr.push(1);}else if(i == 1){arr = new Array();arr.push(1);arr.push(1);}else{var arr2 = new Array();arr2.push(1);for(var j = 0;j<ar