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  • A balanced binary search tree which is also a heap

    - by saeedn
    I'm looking for a data structure where each element in it has two keys. With one of them the structure is a BST and looking at the other one, data structure is a heap. With a little search, I found a structure called Treap. It uses the heap property with a random distribution on heap keys to make the BST balanced! What I want is a Balanced BST, which can be also a heap. The BST in Treap could be unbalanced if I insert elements with heap Key in the order of my choice. Is there such a data structure?

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  • looping through an object (tree)

    - by Val
    Is there a way (in jquery or javascript) to loop through each object and it's children and gandchildren and so on. if so... can i also read their name? example foo :{ bar:'', child:{ grand:{ greatgrand: { //and so on } } } } so the loop should do something like this... loop start if(nameof == 'child'){ //do something } if(nameof == 'bar'){ //do something } if(nameof =='grand'){ //do something } loop end I know this is stupid code but i tried to make it understandable :) btw this is for a jquery UI, as i am clueless how to go on about this. thanks

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  • How to parse time stamps with Unicode characters in Java or Perl?

    - by ram
    I'm trying to make my code as generic as possible. I'm trying to parse install time of a product installation. I will have two files in the product, one that has time stamp I need to parse and other file tells the language of the installation. This is how I'm parsing the timestamp public class ts { public static void main (String[] args){ String installTime = "2009/11/26 \u4e0b\u5348 04:40:54"; //This timestamp I got from the first file. Those unicode charecters are some Chinese charecters...AM/PM I guess //Locale = new Locale();//don't set the language yet SimpleDateFormat df = (SimpleDateFormat)DateFormat.getDateTimeInstance(DateFormat.DEFAULT,DateFormat.DEFAULT); Date instTime = null; try { instTime = df.parse(installTime); } catch (ParseException e) { // TODO Auto-generated catch block e.printStackTrace(); } System.out.println(instTime.toString()); } } The output I get is Parsing Failed java.text.ParseException: Unparseable date: "2009/11/26 \u4e0b\u5348 04:40:54" at java.text.DateFormat.parse(Unknown Source) at ts.main(ts.java:39) Exception in thread "main" java.lang.NullPointerException at ts.main(ts.java:45) It throws exception and at the end when I print it, it shows some proper date... wrong though. I would really appreciate if you could clarify me on these doubts How to parse timestamps that have unicode characters if this is not the proper way? If parsing is failed, how could instTime able to hold some date, wrong though? I know its some chinese,Korean time stamps so I set the locale to zh and ko as follows.. even then same error comes again Locale = new Locale("ko"); Locale = new Locale("ja"); Locale = new Locale("zh"); How can I do the same thing in Perl? I can't use Date::Manip package; Is there any other way?

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  • Error inserting data in binary tree

    - by chepe263
    I copied this code (in spanish) http://www.elrincondelc.com/nuevorincon/index.php?pag=codigos&id=4 and wrote a new one. This is my code: #include <cstdlib> #include <conio.h> #include <iostream> using namespace std; struct nodoarbol { int dato; struct nodoarbol *izq; struct nodoarbol *der; }; typedef nodoarbol Nodo; typedef Nodo *Arbol; void insertar(Arbol *, int); void inorden(Arbol); void postorden(Arbol); void preorden(Arbol); void insertar(Arbol *raiz, int nuevo){ if (*raiz==NULL){ *raiz = (Nodo *)malloc(sizeof(Nodo)); if (*raiz != NULL){ (*raiz)->dato=nuevo; (*raiz)->der=NULL; (*raiz)->izq=NULL; } else{ cout<<"No hay memoria suficiente u ocurrio un error"; } } else{ if (nuevo < (*raiz)->dato) insertar( &((*raiz)->izq), nuevo ); else if (nuevo > (*raiz)->dato) insertar(&((*raiz)->der), nuevo); } }//inseertar void inorden(Arbol raiz){ if (raiz != NULL){ inorden(raiz->izq); cout << raiz->dato << " "; inorden(raiz->der); } } void preorden(Arbol raiz){ if (raiz != NULL){ cout<< raiz->dato << " "; preorden(raiz->izq); preorden(raiz->der); } } void postorden(Arbol raiz){ if (raiz!=NULL){ postorden(raiz->izq); postorden(raiz->der); cout<<raiz->dato<<" "; } } int main() { int i; i=0; int val; Arbol raiz = NULL; for (i=0; i<10; i++){ cout<<"Inserte un numero"; cin>>val; insertar( (raiz), val); } cout<<"\nPreorden\n"; preorden(raiz); cout<<"\nIneorden\n"; inorden(raiz); cout<<"\nPostorden\n"; postorden(raiz); return 0; } I'm using netbeans 7.1.1, mingw32 compiler This is the output: make[2]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' make[1]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' main.cpp: In function 'int main()': main.cpp:110:30: error: cannot convert 'Arbol {aka nodoarbol*}' to 'Nodo** {aka nodoarbol**}' for argument '1' to 'void insertar(Nodo**, int)' make[2]: *** [build/Release/MinGW-Windows/main.o] Error 1 make[1]: *** [.build-conf] Error 2 make: *** [.build-impl] Error 2 BUILD FAILED (exit value 2, total time: 11s) I don't understand what's wrong since i just copied the code (and rewrite it to my own code). I'm really good in php, asp.net (vb) and other languages but c is a headche for me. I've been struggling with this problem for about an hour. Could somebody tell me what could it be?

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  • Binary Search Tree can't delete the root

    - by Ali Zahr
    Everything is working fine in this function, but the problem is that I can't delete the root, I couldn't figure out what's the bug here.I've traced the "else part" it works fine until the return, it returns the old value I don't know why. Plz Help! node *removeNode(node *Root, int key) { node *tmp = new node; if(key > Root->value) Root->right = removeNode(Root->right,key); else if(key < Root->value) Root->left = removeNode(Root->left, key); else if(Root->left != NULL && Root->right != NULL) { node *minNode = findNode(Root->right); Root->value = minNode->value; Root->right = removeNode(Root->right,Root->value); } else { tmp = Root; if(Root->left == NULL) Root = Root->right; else if(Root->right == NULL) Root = Root->left; delete tmp; } return Root; }

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  • mysql category tree search

    - by ffffff
    I have the following schema on MySQL 5.1 CREATE TABLE `mytest` ( `category` varchar(32) , `item_name` varchar(255) KEY `key1` (`category`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1; category column is filled with like that [:parent_parent_cat_id][:parent_cat_id][:leaf_cat_id] "10000200003000" if you can search all of the under categories :parent_parent_category_id SELECT * FROM mytest WHERE category LIKE "10000%"; it's using index key1; but How to use index when I wanna search :parent_cat_id? SELECT * FROM mytest WHERE category LIKE "%20000%"; Do you have a better solutions?

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  • Express XPath as an expression tree

    - by 47d_
    If I have an XPath query like NodeA/NodeB[@WIDTH and not(@WIDTH="20")] | NodeC[@WIDTH and not(@WIDTH="20")]/NodeD Is there any API available to visualize this XPath query as a stack of atomic expressions, something like (following is generic) Get results of NodeA, call it "first set" Get results of NodeB from "first set" Filter where [@WIDTH and not(@WIDTH="20")] Filter NodeD, call this "node d for B" Get results of NodeC from "first set" Filter where [@WIDTH and not(@WIDTH="20")] Filter NodeD, call this "node d for C" Combine "node d for B" and "node d for C" I am trying to see if we can convert the XPath expression into custom expression which is close to english and vice versa. If no API is available, what would be the best approach? Thanks in advance.

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  • transform file/directory structure into 'tree' in javascript

    - by dave
    I have an array of objects that looks like this: [{ name: 'test', size: 0, type: 'directory', path: '/storage/test' }, { name: 'asdf', size: 170, type: 'directory', path: '/storage/test/asdf' }, { name: '2.txt', size: 0, type: 'file', path: '/storage/test/asdf/2.txt' }] There could be any number of arbitrary path's, this is the result of iterating through files and folders within a directory. What I'm trying to do is determine the 'root' node of these. Ultimately, this will be stored in mongodb and use materialized path to determine it's relationships. In this example, /storage/test is a root with no parent. /storage/test/asdf has the parent of /storage/test which is the parent to /storage/test/asdf/2.txt. My question is, how would you go about iterating through this array, to determine the parent's and associated children? Any help in the right direction would be great! Thank you

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  • Javascript / Jquery Tree Travesal question

    - by Copper
    Suppose I have the following <ul> <li>Item 1</li> <li>Item 2 <ul> <li>Sub Item</li> </ul> </li> <li>Item 3</li> </ul> This list is auto-generated by some other code (so adding exclusive id's/class' is out of the question. Suppose I have some jquery code that states that if I mouseover an li, it gets a background color. However, if I mouseover the "Sub Item" list item, "Item 2" will be highlighted as well. How can I make it so that if the user mouses over "Sub Item" it only puts a background color on that and not on "Item 2" as well?

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  • Create Device Reccieve SMS Parse To Text ( SMS Gateway )

    - by Chris Okyen
    I want to use a server as a device to run a script to parse a SMS text in the following way. I. The person types in a specific and special cell phone number (Similar to Facebook’s 32556 number used to post on your wall) II. The user types a text message. III. The user sends the text message. IV. The message is sent to some kind of Device (the server) or SMS Gateway and receives it. V. The thing described above that the message is sent to then parse the test message. I understand that these three question will mix Programming and Server Stuff and could reside here or at DBA.SE How would I make such a cell phone number (described in step I) that would be sent to the Device? How do I create the device that then would receive it? Finally, how do I Parse the text message?

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  • How to find Sub-trees in non-binary tree

    - by kenny
    I have a non-binary tree. I want to find all "sub-trees" that are connected to root. Sub-tree is a a link group of tree nodes. every group is colored in it's own color. What would be be the best approach? Run recursion down and up for every node? The data structure of every treenode is a list of children, list of parents. (the type of children and parents are treenodes) Clarification: Group defined if there is a kind of "closure" between nodes where root itself is not part of the closure. As you can see from the graph you can't travel from pink to other nodes (you CAN NOT use root). From brown node you can travel to it's child so this form another group. Finally you can travel from any cyan node to other cyan nodes so the form another group

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  • GUI question : representing large tree

    - by Peter
    I have a tree-like datastructure of some six levels deep, that I would like to represent on a single webpage (can be tabs, trees; ....) In each level both childnodes and content are possible. Presenting it like a real tree would be not very usable (too big). I was thinking in the lines of hiding parts of the tree when you drill down and presenting a breadcrumbs or the like to keep you informed as to where you are... I guess my question boils down to : any ideas / examples ? Tx!

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  • How to find siblings of a tree?

    - by smallB
    On my interview for an internship, I was asked following question: On a whiteboard write the simplest algorithm with use of recursion which would take a root of a so called binary tree (so called because it is not strictly speaking binary tree) and make every child in this tree connected with its sibling. So if I have: 1 / \ 2 3 / \ \ 4 5 6 / \ 7 8 then the sibling to 2 would be 3, to four five, to five six and to seven eight. I didn't do this, although I was heading in the right direction. Later (next day) at home I did it, but with the use of a debugger. It took me better part of two hours and 50 lines of code. I personally think that this was very difficult question, almost impossible to do correctly on a whiteboard. How would you solve it on a whiteboard? How to apprehend this question without using a debugger?

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  • Quick Outline: Navigating Your PL/SQL Packages in Oracle SQL Developer

    - by thatjeffsmith
    If you’re browsing your packages using the Connections panel, you have a nice tree navigator to click around your packages and your variable, procedure, and functions. Click, click, click all day long, click, click, click while I sing this song… But What if you drill into your PL/SQL source from the worksheet and don’t have the Tree expanded? Let’s say you’re working on your script, something like - Hmm, what goes next again? So I need to reacquaint myself with just what my beer package requires, so I’m going to drill into it by doing a DESCRIBE (via SHIFT+F4), and now I have the package open. The package is open but the tree hasn’t auto-expanded. Please don’t tell me I have to do the click-click-click thing in the tree!?! Just Open the Quick Outline Panel Do you see it? Just right click in the procedure editor – select the ‘Quick Outline’ in the context menu, and voila! The navigational power of the tree, without needing to drill down the tree itself. If I want to drill into my procedure declaration, just click on said procedure name in the Quick Outline panel. This works for both package specs and bodies. Technically you can use this for stand alone procedures and functions, but the real power is demonstrated for packages.

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  • Asymptotic runtime of list-to-tree function

    - by Deestan
    I have a merge function which takes time O(log n) to combine two trees into one, and a listToTree function which converts an initial list of elements to singleton trees and repeatedly calls merge on each successive pair of trees until only one tree remains. Function signatures and relevant implementations are as follows: merge :: Tree a -> Tree a -> Tree a --// O(log n) where n is size of input trees singleton :: a -> Tree a --// O(1) empty :: Tree a --// O(1) listToTree :: [a] -> Tree a --// Supposedly O(n) listToTree = listToTreeR . (map singleton) listToTreeR :: [Tree a] -> Tree a listToTreeR [] = empty listToTreeR (x:[]) = x listToTreeR xs = listToTreeR (mergePairs xs) mergePairs :: [Tree a] -> [Tree a] mergePairs [] = [] mergePairs (x:[]) = [x] mergePairs (x:y:xs) = merge x y : mergePairs xs This is a slightly simplified version of exercise 3.3 in Purely Functional Data Structures by Chris Okasaki. According to the exercise, I shall now show that listToTree takes O(n) time. Which I can't. :-( There are trivially ceil(log n) recursive calls to listToTreeR, meaning ceil(log n) calls to mergePairs. The running time of mergePairs is dependent on the length of the list, and the sizes of the trees. The length of the list is 2^h-1, and the sizes of the trees are log(n/(2^h)), where h=log n is the first recursive step, and h=1 is the last recursive step. Each call to mergePairs thus takes time (2^h-1) * log(n/(2^h)) I'm having trouble taking this analysis any further. Can anyone give me a hint in the right direction?

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  • Binary Search Tree, cannot do traversal

    - by ihm
    Please see BST codes below. It only outputs "5". what did I do wrong? #include <iostream> class bst { public: bst(const int& numb) : root(new node(numb)) {} void insert(const int& numb) { root->insert(new node(numb), root); } void inorder() { root->inorder(root); } private: class node { public: node(const int& numb) : left(NULL), right(NULL) { value = numb; } void insert(node* insertion, node* position) { if (position == NULL) position = insertion; else if (insertion->value > position->value) insert(insertion, position->right); else if (insertion->value < position->value) insert(insertion, position->left); } void inorder(node* tree) { if (tree == NULL) return; inorder(tree->left); std::cout << tree->value << std::endl; inorder(tree->right); } private: node* left; node* right; int value; }; node* root; }; int main() { bst tree(5); tree.insert(4); tree.insert(2); tree.insert(10); tree.insert(14); tree.inorder(); return 0; }

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  • How to cleanly add after-the-fact commits from the same feature into git tree

    - by Dennis
    I am one of two developers on a system. I make most of the commits at this time period. My current git workflow is as such: there is master branch only (no develop/release) I make a new branch when I want to do a feature, do lots of commits, and then when I'm done, I merge that branch back into master, and usually push it to remote. ...except, I am usually not done. I often come back to alter one thing or another and every time I think it is done, but it can be 3-4 commits before I am really done and move onto something else. Problem The problem I have now is that .. my feature branch tree is merged and pushed into master and remote master, and then I realize that I am not really done with that feature, as in I have finishing touches I want to add, where finishing touches may be cosmetic only, or may be significant, but they still belong to that one feature I just worked on. What I do now Currently, when I have extra after-the-fact commits like this, I solve this problem by rolling back my merge, and re-merging my feature branch into master with my new commits, and I do that so that git tree looks clean. One clean feature branch branched out of master and merged back into it. I then push --force my changes to origin, since my origin doesn't see much traffic at the moment, so I can almost count that things will be safe, or I can even talk to other dev if I have to coordinate. But I know it is not a good way to do this in general, as it rewrites what others may have already pulled, causing potential issues. And it did happen even with my dev, where git had to do an extra weird merge when our trees diverged. Other ways to solve this which I deem to be not so great Next best way is to just make those extra commits to the master branch directly, be it fast-forward merge, or not. It doesn't make the tree look as pretty as in my current way I'm solving this, but then it's not rewriting history. Yet another way is to wait. Maybe wait 24 hours and not push things to origin. That way I can rewrite things as I see fit. The con of this approach is time wasted waiting, when people may be waiting for a fix now. Yet another way is to make a "new" feature branch every time I realize I need to fix something extra. I may end up with things like feature-branch feature-branch-html-fix, feature-branch-checkbox-fix, and so on, kind of polluting the git tree somewhat. Is there a way to manage what I am trying to do without the drawbacks I described? I'm going for clean-looking history here, but maybe I need to drop this goal, if technically it is not a possibility.

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  • How can I implement a splay tree that performs the zig operation last, not first?

    - by Jakob
    For my Algorithms & Data Structures class, I've been tasked with implementing a splay tree in Haskell. My algorithm for the splay operation is as follows: If the node to be splayed is the root, the unaltered tree is returned. If the node to be splayed is one level from the root, a zig operation is performed and the resulting tree is returned. If the node to be splayed is two or more levels from the root, a zig-zig or zig-zag operation is performed on the result of splaying the subtree starting at that node, and the resulting tree is returned. This is valid according to my teacher. However, the Wikipedia description of a splay tree says the zig step "will be done only as the last step in a splay operation" whereas in my algorithm it is the first step in a splay operation. I want to implement a splay tree that performs the zig operation last instead of first, but I'm not sure how it would best be done. It seems to me that such an algorithm would become more complex, seeing as how one needs to find the node to be splayed before it can be determined whether a zig operation should be performed or not. How can I implement this in Haskell (or some other functional language)?

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  • How to parse a string to an integer without library functions?

    - by dack
    Hi, I was recently asked this question in an interview: "How could you parse a string of the form '12345' into its integer representation 12345 without using any library functions, and regardless of language?" I thought of two answers, but the interviewer said there was a third. Here are my two solutions: Solution 1: Keep a dictionary which maps '1' = 1, '2' = 2, etc. Then parse the string one character at a time, look up the character in your dictionary, and multiply by place value. Sum the results. Solution 2: Parse the string one character at a time and subtract '0' from each character. This will give you '1' - '0' = 0x1, '2' - '0' = 0x2, etc. Again, multiply by place value and sum the results. Can anyone think of what a third solution might be? Thanks.

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  • Sentence Tree v/s Words List

    - by Rohit Jose
    I was recently tasked with building a Name Entity Recognizer as part of a project. The objective was to parse a given sentence and come up with all the possible combinations of the entities. One approach that was suggested was to keep a lookup table for all the know connector words like articles and conjunctions, remove them from the words list after splitting the sentence on the basis of the spaces. This would leave out the Name Entities in the sentence. A lookup is then done for these identified entities on another lookup table that associates them to the entity type, for example if the sentence was: Remember the Titans was a movie directed by Boaz Yakin, the possible outputs would be: {Remember the Titans,Movie} was {a movie,Movie} directed by {Boaz Yakin,director} {Remember the Titans,Movie} was a movie directed by Boaz Yakin {Remember the Titans,Movie} was {a movie,Movie} directed by Boaz Yakin {Remember the Titans,Movie} was a movie directed by {Boaz Yakin,director} Remember the Titans was {a movie,Movie} directed by Boaz Yakin Remember the Titans was {a movie,Movie} directed by {Boaz Yakin,director} Remember the Titans was a movie directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was a movie directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was a movie directed by {Boaz Yakin,director} The entity lookup table here would contain the following data: Remember the Titans=Movie a movie=Movie Boaz Yakin=director the Titans=Movie the Titans=Sports Team Another alternative logic that was put forward was to build a crude sentence tree that would contain the connector words in the lookup table as parent nodes and do a lookup in the entity table for the leaf node that might contain the entities. The tree that was built for the sentence above would be: The question I am faced with is the benefits of the two approaches, should I be going for the tree approach to represent the sentence parsing, since it provides a more semantic structure? Is there a better approach I should be going for solving it?

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  • vector rotations for branches of a 3d tree

    - by freefallr
    I'm attempting to create a 3d tree procedurally. I'm hoping that someone can check my vector rotation maths, as I'm a bit confused. I'm using an l-system (a recursive algorithm for generating branches). The trunk of the tree is the root node. It's orientation is aligned to the y axis. In the next iteration of the tree (e.g. the first branches), I might create a branch that is oriented say by +10 degrees in the X axis and a similar amount in the Z axis, relative to the trunk. I know that I should keep a rotation matrix at each branch, so that it can be applied to child branches, along with any modifications to the child branch. My questions then: for the trunk, the rotation matrix - is that just the identity matrix * initial orientation vector ? for the first branch (and subsequent branches) - I'll "inherit" the rotation matrix of the parent branch, and apply x and z rotations to that also. e.g. using glm::normalize; using glm::rotateX; using glm::vec4; using glm::mat4; using glm::rotate; vec4 vYAxis = vec4(0.0f, 1.0f, 0.0f, 0.0f); vec4 vInitial = normalize( rotateX( vYAxis, 10.0f ) ); mat4 mRotation = mat4(1.0); // trunk rotation matrix = identity * initial orientation vector mRotation *= vInitial; // first branch = parent rotation matrix * this branches rotations mRotation *= rotate( 10.0f, 1.0f, 0.0f, 0.0f ); // x rotation mRotation *= rotate( 10.0f, 0.0f, 0.0f, 1.0f ); // z rotation Are my maths and approach correct, or am I completely wrong? Finally, I'm using the glm library with OpenGL / C++ for this. Is the order of x rotation and z rotation important?

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