learning javascript data structures and algorithms

Hanoi Problem Solving:Hanoi (also known as Hanoi) is a puzzle toy derived from an ancient Indian legend. When big Brahma created the world, he made three diamond pillars, and stacked 64 gold discs on a pillar from bottom to top in order of size. The great Brahma commanded the Brahman to rearrange the discs from below to the other pillars in order of size. It is also stipulated that the disc cannot be enlarged on the small disc, and only one disc can be moved between the three pillars at a time.

GCD Greatest Common divisorMethod: Euclid Algorithm (Euclidean method),"Thought" recursionIdeasCode1 Public classMain {2 Public Static intgcdintPintq) {3 if(q = 0) {4 returnp;5 }6 intr = p%Q;7 returngcd (q, R);8 }9}LCM Least common multipleThe pursuit of convenient formula method to solve.IdeasCode1 Public classMain {2 Public Static intgcdintPintq) {3 if(q = 0) {4 returnp;5 }6 intr = p%Q;7 ret

Bubble sortThe basic idea of bubble sorting is to compare two adjacent elements each time and swap them out if they are in the wrong order.If there are n numbers to sort, just n?1 the number of digits, which meansN-1 operation. The "Every trip" requires a comparison of two adjacent numbers starting from the 1th position, and the smaller oneIn the back, when the comparison is complete, move back one to continue comparing the size of the two adjacent numbers below, repeat this step until the lastN

; tointk=-1; .intJ=0; - while(JPlen-1) ,{ the //p[k] Represents a prefix, p[j] represents a suffixTenif(k==-1|| P[J]==P[k]) One { A++k; -++J; -Next[J]=k; the} -Else -{ -k=Next[k]; +} -} +}The value of Next[j] (that is, K) indicates the next move position of the J pointer when p[j]! = T[i].When J is 0 o'clock, if this time does not match, J already on the leftmost, can not move again, this time should be I pointer back move. So in the code there will be next[0] =-1; this initialization.P[K]! = P[

① problem: If there is a tuple or sequence containing n multiple elements, now you want to break it down into n separate variables. 1 L = (4, 5)2 x, y = lView Code② Advanced article:data = ['sb' ' big hammer ', (2018, 6, 2 = dataa='sb'b=' big hammer ' C=51Date= (2018, 6, 2)③data = ['sb' ' big hammer ', (2018, 6, 2 == 2018= 6= 2View CodeSo then the question comes, this is to know the list of how much of the situation to achieve, if the unknown? that

1. Code implementationdefMerge_sort (alist):ifLen (alist) : returnalist#two-part decompositionnum = Len (alist)/2 Left=Merge_sort (Alist[:num]) right=Merge_sort (alist[num:])#Merging returnmerge (Left,right)defmerge (left, right):" "merge operation, combine two ordered arrays left[] and right[] into a large ordered array" " #The subscript pointer to left and rightL, R =0, 0 result= [] whileL andrLen (right):ifLEFT[L] Right[r]: Result.append (Left[l]) L+ = 1Else: Result.append (R

1. How to find two pointsFind the fastest speed for a sorted list2. Code implementation(1) Recursive implementationdefBinary_search (Alist, item):ifLen (alist) = =0:returnFalseElse: Midpoint= Len (alist)//2ifalist[midpoint]==Item:returnTrueElse: ifitemAlist[midpoint]:returnBinary_search (Alist[:midpoint],item)Else: returnBinary_search (alist[midpoint+1:],item) testlist= [0, 1, 2, 8, 13, 17, 19, 32, 42,]Print(Binary_search (Testlist, 3))Print(Binary_search (Testlist, 13))(2) N

1. Principle:2. Code implementationdefQuick_sort (alist, Start, end):"""Quick Sort""" #Recursive exit conditions ifStart >=End:return #sets the starting element as the datum element to find the positionMID =Alist[start]#Low is the left-to-right cursor for the left of the sequenceLow =Start#High is the right-to-left cursor to the right of the sequenceHigh =End whileLow High :#If low is not coincident with high, the element with high points is not smaller than the datum element, then high

1. Basic properties of stacks and queues Stack is advanced after out; (like a bullet clip, a last-shot first) Queues are FIFO; (like waiting in line to buy ice cream, in order to take turns) Stack and queue in the implementation of the structure can have an array and a list of two forms; (1) The array structure is easy to implement;(2) The structure of the linked list is more complicated, because it involves a lot of pointer operation;1.1 Basic operation of stack structure(1) Pop op

()) { CharCH =Thestack.pop (); Output= output +ch; } returnoutput; } } classReverseapp { Public Static voidMain (string[] args)throwsIOException {String input, output; while(true) {System.out.print ("Enter A string:"); System.out.flush (); Input=getString1 (); if(Input.equals ("")) { Break; } reverser Thereverser=Newreverser (input); Output=Thereverser.dorev (); System.out.println ("Reversed:" +output); } } Public StaticString getString

This article illustrates the fast sequencing of data structures and algorithms implemented by Python. Share to everyone for your reference. The specific analysis is as follows: I. Overview Quick Sort is a sort of divide-and-conquer algorithm. The algorithm first selects a partitioning element (partition element, sometimes called a pivot), and then rearranges th

than D), but the cardinality sort is only suitable for a key code with distinct structural characteristics such as strings and integers. When n is large and d is small, it can be sorted by cardinality. (5The sorting algorithm discussed earlier, in addition to the Cardinal sort, the other sorting algorithm is the sequential storage structure. In order to avoid a significant amount of time to record movement, a chain storage structure can be used in order to sort the records. Direct Insert sor

data in the stack:int top;initially, when the stack is empty, set: top =–1to avoid stack overflow, you need to check if the stack is full before adding elements to the stack. Let's modify this algorithm to check for this condition. Problem Description:write a program to implement a stack by using an array that requires the stack to hold5an element. So-kinsoku-overflow:1 ' >Using system;using system.collections.generic;using system.componentmodel;usin

all cases:650) this.width=650; "src=" Http://files.jb51.net/file_images/article/201211/2012110120523847.png "/>void Clear (); empty queue; Initial conditions: queue exists; operation result: Leave the queue empty.void in (T item); Queued Initial conditions: queue exists; action Result: Adds a new data element with a value of item to the end of the queue, and the queues are changed.650) this.width=650; "Width=" 620 "height=" 437 "src=" http://files.jb

Binary tree node:/** Binary tree node*/public class Node {Data itempublic Long data;Left Dial Hand nodePublic Node Leftchild;Right child nodePublic Node Rightchild;/*** Construction Method* @param data*/Public Node (Long data) {This.data = data;}}Binary Tree class:/** Two-fo

(recorded as n) should be equal to the sum of 0 degree nodes, 1 degree nodes (recorded as N1) and 2 degree nodes:N=NO+N1+N2 (Formula 1)On the other hand, 1-degree nodes have a child, 2-degree nodes have two children, so the total number of child nodes in the binary tree is:Nl+2n2Only the root node in the tree is not a child of any node, so the total number of nodes in the binary tree can be expressed as:N=n1+2n2+1 (Formula 2)obtained by the Formula 1 and the 2:No=n2+1"

= mid#determine the starting subscript for the post-half string - Else: -Next = mid + 1 the -top = Mid-1 - forIinchRange (Next,len (A)): - ifStack[top]! =A[i]: + returnFalse -Top-= 1 + A returnTrueTest:if __name__= ="__main__" := List ("Hahahahahahahahaha ") print is_plalindrome_demo1 (q)Manual implementation Stack1 #simple filo stack category2 classStack:3 def __init__(self):4Self.top = None#point to top of stack5Self.end = None#Point to the bo

field to delete*/Public Node Delete (Long value) {Node current = first;Node previous = first;while (current.data! = value) {if (Current.next = = null) {return null;}previous = current;current = Current.next;}if (current = = first) {first = First.next;} else {Previous.next = Current.next;}return current;}}Test:public class Testlinklist {public static void Main (string[] args) {linklist linklist = new linklist ();Linklist.insertfirst (34);Linklist.insertfirst (23);Linklist.insertfirst (12);Linkli

string into the object.Remove removes the character from the StringBuilder object. Two parameters: The starting position and the number of characters to remove. ‘Replace: Replaces the character of the StringBuilder object. Two parameters: The old string to replace and the new string to replace.When working with StringBuilder objects, it is often necessary to convert them into strings, which can be implemented using the ToString method. Returns a string instance in the StringBuilder instance.C #

the first chapter: Data structure and algorithmIntroduction: This book is advanced usage, not small white use bookPurpose: The purpose of writing is to record the process of learning the book and the Harvest1.1 The decompression sequence is assigned to multiple variables:problem : There is now one containing n elements of a tuple or a sequence, how to extract the value inside it and assign a value to n vari