best book for data structures and algorithms in java

Quick sort:/** Quick Sort*/public class QuickSort {/*** Divide arrays*/public static int partition (long arr[],int left, int. Right,long point) {int leftptr = left-1;int rightptr = right;while (true) {Loop, will be smaller than the keyword left in the left sidewhile (Leftptr Loop, will be larger than the keyword left on the right sidewhile (Rightptr > Leftptr arr[--rightptr] > point);if (leftptr >= rightptr) {Break} else {Long tmp = arr[leftptr];ARR[LEFTPTR] = arr[rightptr];ARR[RIGHTPTR] = tmp;

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

1#include 2#include 3 4 intBinsearch (intXintV[],intn);//function Declaration5 6 intMain ()7 {8 intarr[]={1,2,3,4,5,6,7,8,9,Ten};9 intas a result, num;Tennum=7; OneResults =binsearch (Num,arr,Ten); A - if(Results 0) -printf"I didn't find!\n ."); the Else -printf"find%d\n in arr[%d]", as a result, num); -System"Pause"); - + return 0; - } + A intBinsearch (intXintV[],intN//function Definition at { - intLow,high,mid;//min, max, Middle - -low=0; -high=n-1; - in

Data structures and algorithms (C # implementation) series --- AVLTree (II) // --------------- Override -------------------- Public override void AttachKey (object _ obj) { If (! IsEmpty ()) Throw new Exception ("My: this node must be a empty tree node! "); This. key = _ obj;

) { returnNULL; } TreeNode* node =NewTreeNode;//noice that [new] should is written out.Node->elem = * (aftorder+length-1); Std::coutStd::endl; intRootindex = 0; for(; rootindex //a variation of the loop { if(Inorder[rootindex] = = * (aftorder+length-1)) Break; } node->left =binarytreefromorderings (inorder, Aftorder, Rootindex); Node->right = binarytreefromorderings (inorder + rootindex + 1, Aftorder + rootindex, Length-(Rootindex + 1)); returnnode;}intMainintarg

Data structures and algorithms (C # implementation) series --- tree (III) Heavenkiller (original) // Overwrite Object. Equals () --- reference type realization Public override bool Equals (object _ obj) { If (_ obj = null) Return false; // this cannot be null If (! (This. GetType () = _ obj. GetType ())) Return false; // The type is not equal. Tree tmpObj = (Tre

Public classSelectionsort { Public Static voidsort (integer[] array) {if(Array = =NULL|| Array.Length = = 0) { return; } Integer mininum=NULL; Integer Miniindex=NULL; for(inti = 0; i ) {Miniindex=i; Mininum=Array[i]; for(intj = i + 1; J ) { if(Array[j] mininum) {Miniindex=J; Mininum=Array[j]; } } if(Miniindex! =i) {swap (array, I, miniindex); } } } Private Static voidswap (integer[] array, integer i, Integer j) {Integer temp=Array[i]

1 Sequence decompression: through * to pass the match*a, B = somelist, First, *mid, last = Somelist, a, *b = Somelist2 using bidirectional queues: From collections import dequeQ = deque (maxlen=5) can be fixed lengthQ = deque () can also be any lengthcan be inserted and removed from both ends, append, Appendleft, pop, popleft3 finding the largest or smallest n elements: using heapq (heap queue)HEAPQ. nlargest (N, Alist) heapq. nsmallest (n, alist) is suitable when n is relatively smallYou can al

The maximum value in the output array A and its subscript#include Data structures and algorithms-maximum values in a string output array

Title: Define the Fibonacci sequence as follows:/0 N=0F (n) = 1 n=1\ f (n-1) +f (n-2) n=2Enter N to find the nth item of the sequence in the quickest way.#include Data structures and algorithms-string Fibonacci to find the nth item

(self) self.engine.accept (visitor) self.body.accept (Visito R) forWheelinchself.wheels:wheel.accept (visitor)#This is our visitor, and every change is here .classPrintvisitor:defVisitwheel (self, Wheel):Print "Visiting"+wheel.name+"Wheel" defVisitengine (self, Engine):Print "Visiting engine" defvisitbody (self, body):Print "Visiting Body" defVisitcar (self, car):Print "Visiting Car"if __name__=='__main__': Car=Car () visitor=printvisitor () car.accept (visitor)Learn some Python basics

Package Com.js.ai.modules.pointwall.testxfz;class ordarray{private long[] a;private int nelems;public OrdArray (int max ) {a=new long[max];nelems=0;} public int size () {return nelems;} Insert method public void Insert (Long value) {int j;for (j=0;j　　Data structures and algorithms: the dichotomy Demo

True Second, collections. Ordereddict: Dictionary with OrderOrdereddict internally maintains a doubly linked list to record the order in which key values are inserted, update the key value does not affect the original order, insert key value is inserted at the end, so its memory consumption is twice times the normal dictionary ,From collections Import ORDEREDDICTD = Ordereddict () d[' a '] = 1d[' c '] = 3d[' b '] = 2print (d, D.items ()) Ordereddict (' A ', 1), (' C ', 3)

1. The principle of hill sorting2. Code implementationdefShell_sort (alist): N=Len (alist)#Initial StepGap = N/2 whileGap >0:#Insert Sort by step forIinchRange (Gap, N): J=I#Insert Sort whileJ>=gap andALIST[J-GAP] >Alist[j]: alist[j-GAP], alist[j] = alist[j], alist[j-Gap] J-=Gap#get a new step sizeGap = GAP/2alist= [54,26,93,17,77,31,44,55,20]shell_sort (alist)Print(alist)Attention:(1) Hill sort uses the idea of splitting up and merging, splitting apart3. Complexity of TimeOp

node is now Point to the head of the now node's next pointer; Set the now node to the head of the new rollover completion node header now; Point to head now, and so on, the next pointer to the previous node of the now node. 3. Two points search common scenes Finding a number in an ordered sequence; For example, given an array of arr, determine if the integer m is in arr (Train of thought: Determine the size of the mid-to-m re

The implementation of the dictionary// 字典类function Dictionary () { this.add = add; this.dataStore = new Array(); this.find = find; this.remove = remove; this.showAll = showAll; this.count = count; this.clear = clear;}function add (key, value) { this.dataStore[key] = value;}function find (key) { return this.dataStore[key];}function remove (key) { delete this.dataStore[key];}function showAll () { let datakeys = Array.prototype.slice.call(Object.keys(this.dataStore)

node1.next = node3; node3.previous = node1; node3.next = node2; node2.previous = node3; node2.next = node4; node4.previous = node2; }};DList.prototype.advance = function (n, item) { let currNode = this.find(item); while (n--) { this.goPrevious(currNode); }};// 示例let names = new DList();names.insert('Mazey', 'head');names.insert('Cherrie', 'Mazey');names.insert('John', 'Cherrie');names.insert('Luna', 'John');names.insert('Ada', 'Luna'

efficiency: In a merge sort, the maximum value of this secondary storage space is not more than N, so the spatial complexity of the merge algorithm is θ (n) in order to combine the sub-sequences to use additional storage space.Time efficiency: The merging algorithm is a typical divide-and-conquer algorithm with a time complexity of 0 (n log n).　3. Algorithm examplesMergesort.javaPackage Com.test.sort.merge;public class MergeSort {/** * @param args */public static void main (string[] args) {//TO

" definition of algorithmic time complexity"At the time of the algorithm analysis, the total number of executions of the statement T (N) is a function of the problem size n, which then analyzes the change of T (n) with N and determines the order of magnitude of T (N). The time complexity of the algorithm, which is the time measurement of the algorithm, is recorded as: T (n) = O (f (n)). It indicates that with the increase of the problem size n, the growth rate of the algorithm execution time is

and J to point to the leftmost and rightmost of the sequence, respectively. At first, the variable I points to the leftmost part of the sequence, which is the number 6, and J points to the far right of the sequence, pointing to the number 8, as shown inFirst J moves from right to left, so long as it finds a number smaller than 6, it stops, and J is in the 5 position. I then start from left to right, as long as I find a number greater than 6 to stop, so it to the position of 7This is where the t