good book for data structures and algorithms

, vi belongs to V-S. Then, the shortest path is compared with the source point A to the path length of the node vi, the minimum length is the shortest path from source point A to Node VI. Finally, the end of the shortest path (i.e. VI) is added to the S(3) Repeat step (2) until the shortest path of all endpoints is found, that is, the number of nodes in S is equal to the number of nodes in V.The time complexity of the Dijkstra algorithm is O (n^2), the spatial complexity depends on the storage m

Data structures and algorithms (C # implementation) series-demonstration (II) Heavenkiller (original) Public static void ShowGeneralTree_travel () { IEnumerator tmpIEnum; Tree. TraversalType travelType = 0; // --------------------- Prompt ---------------------------- Console. WriteLine ("please choose a No. of a item you want to travel :"); Console. WriteLine ("

Data structures and algorithms (C # implementation) series --- tree (2) Heavenkiller (original) Public class InOrder: IPrePostVisitor { Private IVisitor visitor; Public InOrder (IVisitor _ vis) {visitor = _ vis ;} # Region IPrePostVisitor Member Public void PreVisit (object _ obj) { // TODO: Add InOrder. PreVisit implementation } Public void Visit (object _ ob

Data structures and algorithms (C # implementation) series --- generalized tree (II) Heavenkiller (original) Public override object Key {get {return this. key ;}} Public override uint Degree {get {return this. degree ;}} // Public override uint Height {get {return this. height ;}} Public override bool IsEmpty () // property takes the place of IsEmpty () {Return

It seems that the following is the most elegant implementation.Other, either node redundancy, or initialize ugly ...#!/usr/bin/env python#-*-coding:utf-8-*-classNode:def __init__(self, initdata): Self.__data=InitData self.__next=NonedefGetData (self):returnSelf.__data defGetNext (self):returnSelf.__next defSetData (Self, newdata): Self.__data=NewDatadefSetnext (Self, newnext): Self.__next=Newnextclasssincyclinkedlist:def __init__(self): Self.head=Node (None) self.head.setNext (self.head)defAdd

next time . A the defShufflepile (self): + """in the current state, the tree is adjusted so that it becomes a heap""" - #downward adjustment from "heap bottom" to "heap top", which keeps the smallest elements rising $ #This allows the heap below the I node to be the local minimum heap. $ forIinchRange ((Len (self)-2)/2,-1,-1):#N/2,..., 0 - Self.siftdown (i) - the defdeletemin (self): - """Remove Minimum element"""Wuyit = self[0]#record the

//Delete a node in a binary sort tree the PublicBstree Deletebstree (Bstree bstree,intkey) { the if(Bstree = =NULL) { - return NULL; - } the the if(Bstree.data = =key) { the //first case: leaf node the if(Bstree.left = =NULL Bstree.right = =NULL) { -Bstree =NULL; the } the //second case: node has left dial hand nodes. the if(Bstree.left! =NULL Bstree.right = =NULL) {94Bstree =Bstree.left; t

1 PackageIYou.neugle.search;2 3 Public classHash_search {4 Private Static intm = 13;5 Private Static int[] hash =New int[m];6 Private Static int[] Array =New int[] {13, 25, 2, 60, 39, 52 };7 8 Public Static voidMain (string[] args) {9 Inserthash ();TenSYSTEM.OUT.PRINTLN ("Hash table is established as follows:"); OneSystem.out.print ("["); A for(inti = 0; i ) { - System.out.print (Hash[i]); - if(I! = hash.length-1) { theSystem.out.print (","); -

　　1 //input A, B, output a+b2 /*#include 3 int main ()4 {5 int A, b;6 scanf ("%d%d", a,b);7 printf ("%d", a+b);8 }*/9 /*Ten //Enter a character to return his ASCLL code One #include A int main () - { - char A; the scanf ("%c", a); - printf ("%d", a); - }*/ - + -#include + intMain () A { at CharA; - Charb; - while(scanf ("%c%c", a,b)!=eof)//when there is a space in the middle, the input parameters should also have a space, when there is no space, you do not have to enter a space.

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.

Stack: LIFO. Stack top at the end, bottom of the stack in the front. The newly added element and the element to be deleted are saved at the end of the stack.Create a stack:functionStack () {varitems = [];/*Save the elements in the stack with an array*/ This. Push =function(e) {Items.push (e); } This. Pop =function() { returnItems.pop (); } This. Peek =function() { returnItems[length-1]; } This. IsEmpty =function() { returnItems.length = = 0; } This. Size

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

only inherit one parent classPolymorphicFour, algorithm effectiveness analysis:4.1 Time Complexity: Https://baike.baidu.com/item/%E6%97%B6%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6/1894057?fr=aladdin (Baidu Connection)　　F (N) =n! (factorial) n-times with a time complexity of n4.2 Space Complexity: Https://baike.baidu.com/item/%E7%A9%BA%E9%97%B4%E5%A4%8D%E6%9D%82%E5%BA%A6/9664257?fr=aladdin (Baidu Connection)V. Abstract classTo include one or more abstract methods (without a method body): To implement

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

numbering unit 　　You can use the stack to do 10-in-2~9 binary operationsHere's how: a decimal number A, binary B1> will a%b, press into the stack2> Replace A with A/b3> if a is greater than 0, continue to hit 1> repeatedlyIf it is less than 0, jump out4> the elements of the stack pop up once, forming a new character (the character is the result of conversion completion)As an example:　　10 to 2 binary:10%2 = 0--into the stack--05%2 =--1 in the stack, 02%2 = 0--into the stack--0, 1, 01%2 =--1 in s

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