data structures udemy

=currentlink.next;} System.out.println (""); System.out.print ("Linklist.displaylist () Finish");}}public class Linklistapp {public static void main (string[] args) {linklist thelinklist =new linklist (); Thelinklist.insertlink (9, 999, 999.999); Thelinklist.displaylist (); Thelinklist.insertfirst (22, 22.22); Thelinklist.insertfirst (33.33), Thelinklist.insertfirst (44.44), Thelinklist.insertfirst (55, 55.55); Thelinklist.displaylist (); Thelinklist.insertlink (2, 333, 333.333); Thelinklist.di

Have you ever worried about "Searching for Matching content in complicated JSON data structures"? This article introduces eight ways to query json data structures, there is always a method suitable for your projectThe following eight methods can be used to query json data

JSON is a lightweight, text-based data interchange format that can easily pass character information in a specific character format, with a detailed interpretation of JSON below.Json:Full name: "JavaScript Object Notation", a text-based, language-independent, lightweight data Interchange format, in my opinion JSON features are concise, useful, in all the languages I have access to, Can be used across platfo

Set and criptset of JavaScript data structures and algorithms Set) When talking about collections, I think that when I first got into high school, the first class of mathematics was about collections. Therefore, I feel more cordial when learning the data structure of collection.There is one basic property of a set: Elements in the set are not repeated. Because of

); Linkedlist.removefirst (); Linkedlist.removelast (); //determine if the linked list is emptySystem.out.println (Linkedlist.isempty ());} HashMap Non-synchronous Public Static voidhashmaptest () {HashMapNewHashmap(); //adding elements to the hash tableHashmap.put ("TP", 123); Hashmap.put ("ZP", 567); System.out.println (HASHMAP); //facilitates a hash table with iterators for(IteratorHashmap.entryset (). iterator (); Iterator.hasnext ();) {EntryIterator.next (); Syst

. listsize; return This. listsize; }; This. Find =function(ele) { for(vari=0; i This. listsize; i++){ if( This. datastore[i] = = =ele) { returni; } } return-1; }; This. add=function(ele) { This. datastore[ This. listsize++] =Ele; }; This. Insert =function(Ele, Afterele) {varindex = This. Find (Afterele); if(Index >-1){ This. Datastore.splice (Index, 0, ele); This. listsize++; return true; }Else{ return false; } };

top element and adjust r[1..i-1] to new heap int temp = r[1];//Exchange heap top and base element r[1] = R[i];r[i] = Temp;heapadjust (r, 1, i-1);//Adjustment}}//has Know R[low. High] except R[low], the rest of the elements satisfy the definition of the heap private void Heapadjust (int[] r, int low, int.) {int temp = r[low];for (int j = 2 * low; J 　　C: algorithm ExampleHeapsort.javaPackage Com.test.sort.selection;public class Heapsort {/** * @param args */public static void main (string[] args)

1.mapMap2.enum (enumeration)Enum keyword{first,second,third}void func (keyword key) { switch (key) {case First :; Case SECOND:; } }3. Arraysint numbers[2];//numeric size is constant i = f ();vector　　C + + Fundamentals (i): Data types and structures

element, up to log (n) ancestors.So the modification algorithm is as follows (to a node I plus x):Step1: When i > N, the algorithm ends, otherwise turns the second step;Step2:ci = Ci + x, i = i + lowbit (i) go to the first step.i = i +lowbit (i) This process is actually just a process of filling the end 1 with 0.The tree array is too fast for the sum of the arrays!Note:Suggested formula for Lowbit (x):Lowbit (x): =x and (x xor (x-1));or Lowbit (x): =x and (-X);The following question is attached

algorithm is executed on different computers, and the execution time is not the same.Therefore, the operational efficiency should be related to the execution steps, which will be the time complexity of the execution steps.In the first algorithm: T (n) = n^3 * 2In the second algorithm: T (n) = n^2 * 3If the system and bias items are not considered, then the progressive function, using the progressive function representation, is the large O notation:In the first algorithm: T (n) = O (n^3)In the s

));}} //Look at cardsLookpoker ("Wind Qing", Fengqingyang);Lookpoker ("Brigitte", Linqingxia);Lookpoker ("Elina", Liuyi); lookpoker ("hole cards", Dipai);}public static void Lookpoker (String name, arraylistSystem.out.print (name + "The card is:");for (String S:array) {System.out.print (S + "");}System.out.println ();}} Test multiple timesOutput: The wind of the card is: Xiao Wang ♦8♥10♦5♣q♥4♣7♣8♠9♦a♠4♥7♠2♦6♠7♥8♣9Brigitte's cards are: ♦q♥5♥9♠a♠6♣3♠q♦7♦10♠k♦k♣k♥3♠10♠8♥6♠5Elina's cards are: ♥2♠3♥

(Array.get (x)); } NBSP;NBSP;} Look at the cardsLookpoker ("Wind Qing", Fengqingyang);Lookpoker ("Brigitte", Linqingxia);Lookpoker ("Elina", Liuyi); lookpoker ("hole cards", Dipai);}public static void Lookpoker (String name, arraylistSystem.out.print (name + "The card is:");for (String S:array) {System.out.print (S + "");}System.out.println ();}} Test multiple timesOutput: The wind is the card is: Xiao Wang 8? 10? 5? Q. 4? 7? 8? 9? A 4? 7 2? 6? 7? 8? 9Brigitte's cards are:? Q. 5? 9? A? 6? 3?

Author : DS Course GroupUnit : Zhejiang University2-18the adjacency table for a given graph is as follows. Starting from vertex V1 by the breadth-first search method, the resulting vertex sequence is: (2 points) V1,v2,v3,v4,v5 V1,v2,v3,v5,v4 V1,v3,v2,v4,v5 V1,v4,v3,v5,v2 Author : DS Course GroupUnit : Zhejiang University2-19The adjacency matrix of a graph is known as follows, from Vertex V1 start by the breadth-first search method, a vertex sequence that may be ob

#include "stdio.h"#include "Stdlib.h"void swap (int *a,int *b)//Exchange two numbers{int t;T=*a;*a=*b;*b=t;}void minheap (int *a,int i)//Keng Gen adjustment, in the I-node upward adjustment, based on a given array, the creation of small Gan for the arrays{int temp,j;Temp=a[i];j= (i-1)/2; The parent node of the I nodewhile (j>=0i!=0){if (a[j]A[I]=A[J];I=j;j= (i-1)/2;}A[i]=temp;}void Add (int *a,int key,int N)//insert a node at the end and make an adjustment{A[n]=key;Minheap (A,n);}void Minheapfix

the character before and after the pointer.while (p>s) { *p=*p^*s; *s=*p^*S; *p=*p^*S; p--; s++;} Question 3: How to find the first occurrence of a character in a stringMethod 1: Use the hash table to record the number of occurrences of each character, with the character as the array subscript, and the array size to 256. The first time the scan element will be character statistics, the second pass through to find the first count of 1 characters.Task 4: All subsets of a stringRef

, intersection, and difference operations. So, if you want to perform some normal collection operations on the keys of the collection, you can use the key view objects directly without first converting them to a set.The dictionary's items () method returns an element view object that contains (key, value) pairs. This object also supports collection operations and can be used to find out which two dictionaries have the same key-value pairs.Although the values () method of the dictionary is simila

, and clear methods for that collection). But if the iterator calls its own remove method, it is still legal.3.List interfaceThe Java.util package inherits the collection interface. Here are just a few important methods. Public Interface extends Collection{ AnyType get (int idx); AnyType Set (int idx, AnyType newval); void Add (int idx, AnyType x); void Remove (int idx); Listiteratorint poss);}View CodeIndex 0 is in the front segment of the table, size ()-1 represents t

Tag:prethread Array readinggen thought namelog time Using system;using system.collections.generic;using system.linq;using system.text;using System.Threading.Tasks; Namespace Insert Sort {class Program {static void Main (string[] args) {int[] Arry = {10,3,4,6,7 , 9,8,2,1,0}; Insertsortpro (Arry, 10); for (int i=0;i"Normal insert Sort" is slower than selecting sort, but optimized insert ordering is faster than selecting sort. Insert sort has a feature when the ar

("pause");}void Push (Stack s){Elemtype e;Int J, N;if (s.stacksize = = 0){cout System ("pause");Return}if (s.top = = s.stacksize){cout System ("pause");Return}cout CIN >> N;cout for (j = 0; J {CIN >> E;* (S.base + s.top) = e; Insert a new element++s.top; Stack top pointer moves back}System ("pause");}void Pop (Stack s){Elemtype e;Int J, N;if (s.stacksize = = 0){cout System ("pause");Return}if (S.top = = 0){cout System ("pause");Return}cout CIN >> N;cout for (j = 0; J {E = * (S.base + s.top-1);

}return k;}void DFS (mgraph G, int v){Int J;cout VISITED[V] = TRUE;for (j = 0; J {if (G.kind = = DG | | G.kind = = AG)if (!visited[j] G.arcs[v][j].adj = = 1)DFS (G, J);if (G.kind = = DN | | G.kind = = an)if (!visited[j] G.arcs[v][j].adj! = INFINITY)DFS (G, J);}}void Dfstraverse (Mgraph G){int V;for (v = 0; v VISITED[V] = FALSE; Initialize to False, and change to true after traversalfor (v = 0; v if (!visited[v])DFS (G, v);} Void Bfstraverse (Mgraph g) {//The breadth-first search for the graph