data structures and algorithms in java 2nd edition

//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 (","); -

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

()) { 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

-array with only one data item, and the other sub-array containing N-1 data items.A: in this case, the benefits of partitioning are gone, and the execution efficiency of the algorithm is reduced to O (N2). In addition to being slow, there is another potential problem, and when the number of partitions increases, the number of calls to the recursive method increases, and each method call increases the size o

(e e);//replace element in collection * "9" void Add (e e);//At the previous position of the current index or the current Adds a new element to the collection at the back of the index position. *//whether or not to add new elements before or after the current index position depends on whether you are traversing sequentially or in reverse order * If it is {@link#next}, insert at the previous position of the current position *//if it is {@link#previous} to inser

]; while (end>start) { while (end>startarrys[key)//If there is nothing smaller than the key value, compare next until there is a smaller swap position than the key value, And then compare it back to the past End--; if (arrys[end]) { arrys[end]; arrys[arrys[start]; arrys[tmp; } while (end>startarrys[start]) Start+ +; if (arrys[key) {

1 and 0 occurrences of the place, 1 and 0 appear two times, since there is no 0, then 1 is occupy the first and second position (please read this sentence carefully, this sentence read, the whole algorithm understand). Then we put one of the 1 in the position of the second position of the 1 b[1], while c[1]-1, because we have already lined up a 1.Next, I=1,a[1]=2,c[2] is less than or equal to 2 of the number of occurrences, c[2]=4, then to rank it in fourth place, that is b[3] position, while c

(String)); }Second, the use of the stack-delimiter matching.public int charat (char c) {for (int i = 0; i will ({Press into the stack once, once encountered)}] will be compared with the popup element, if it matches, then match. If not)}], the left sign of the stack pops up and the hint is where the specific right symbol type is missing. This is a tool that can be implemented using stacks.Stack of data into the stack and the time complexity of the

of red-Haishi. But they can't all be full, so the height of the 2-3-4 tree is roughly log2 (n+1) and log2 (n+1)/2. Reducing the height of the 2-3-4 tree makes it shorter to find than red-Haishi.On the other hand, each node has more data items to view, which increases the lookup time. Because a linear search is used in a node to view data items, the multiple of the lookup time is proportional to M, which is

1. HashMap Overview HashMap is an asynchronous implementation of the map interface based on a hash table (Hashtable is similar to HashMap, the only difference being that the method in Hashtalbe is thread-safe, that is, synchronous). This implementation provides all the optional mapping operations and allows NULL values and NULL keys to be used. This class does not guarantee the order of mappings, especially if it does not guarantee that the order is immutable. 2. Four points of attention on Has

starts with two pointers, and two pointers point to both ends of the array. (The word "pointer" is used here to indicate the array data item, not the pointer in C + +.) The pointer on the left, Leftptr, moves to the right, while the pointer on the right, rightptr, moves to the left.In fact, leftptr initialization is at the left of the first data item, and Rightptr is on the right-hand side of the last

, while (!thestack.isempty ()) {int Currentvertex = Thestack.peek (); int v = Getadjunvisitedvertex (Currentvertex); if (v = =-1) {Thestack.pop ();} else{vertexlist[v].wasvisited = True;thestack.push (v);d Isplayvertex (Currentvertex);d Isplayvertex (v); System.out.print ("");}} Search complete, initialize, for the next search for (int i = 0; i 5. SummaryThe graph is made up of vertices connected by edges, which can represent many real world situations, including aircraft routes, electronic circ

can put in *, in Put +, that is abc*+.A * (b+c) turn suffix:Read a, read *, later found (can not be placed *, (B+C * Priority High, read, followed by), can be changed to bc+, and a * together, that is abc+*.This is a queue, but the structure to use is the stack, because it comes in. The arithmetic here does not have to be pressed into the stack, directly output.a+b-c Turn suffix:Direct output A, read +, stack empty press into the stack, then output B, stack non-empty, pop +, read to-,+ priority

Preface to the new book "algorithms-principles hidden behind data structures", data structures and algorithms Preface to the book "algorithms-principles hidden behind

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JavaScript data structures and algorithms stack and queue, data structures and algorithms Learning cause Once upon a visit to V2EX, I encountered such a post.The mathematics is completely returned to the teacher. I want to learn s

solving methods and expect to be confident in your ability to deal with problems, and you know the difficulty of learning some programs. The complexity of large-scale problems and the corresponding solutions greatly outweigh the underlying issues associated with the approach. This chapter focuses on two aspects of the content. First, review the frameworks that must be followed in the study of data structures