chouclee

%% Iterative SVD 
%  Calculate the largetst singular value and it's corrsponding
% left-singular vector and right-singular vector
% Author: chouclee
% Date: 10/01/2014
function [L,w,R] = SVD(S)
% S=[6,8,5,9;
%     8,5,4,6;
%     9,8,7,7;
%     6,7,3,8;

chouclee

// if current node has right subtree, return most left node of the right child.
// else trace back to its parent, if the parent is smaller than current node, then consider parent's parent...
// until parent node is bigger than current node.
public class findNextNode<Key extends Comparable<Key>> {
	private Node root;
	
	private class Node {
		private Key key;
		private Node left, right;
		private Node parent;

chouclee

public class Node<Key> {
	public Key key;
	public Node<Key> left, right;
	
	public Node(Key key) {
		this.key = key;
	}
}

chouclee

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.ListIterator;

public class BST<Key extends Comparable<Key>> {
	private Node root;
	
	private class Node {
		private Key key;
		private Node left, right;

chouclee

import java.util.LinkedList;

public class BST<Key extends Comparable<Key>> {
	private Node root;
	
	private class Node {
		Key key;
		Node left, right;
		
		public Node(Key key) {

chouclee

/* Depth First Search	*/

import java.util.ArrayList;

public class DGraph {
	private int V;
	private ArrayList<Integer>[] adj;
	private boolean[] marked;
	
	@SuppressWarnings("unchecked")

chouclee

public class BST<Key extends Comparable<Key>> {
	private Node root;
	private boolean isBalanced;
	
	private class Node {
		private Key key;
		private Node left, right;
		
		public Node(Key key) 

chouclee

public class BST<Key extends Comparable<Key>> {
	private Node root;
	
	private class Node {
		private Key key;
		private Node left, right;
		
		public Node(Key key) 
		{ this.key = key; }

chouclee

// if stack.size() is not allowed, use insertion sort, but it's an O(n^2) (time) algorithm, space : O(n)
// else, we can use MergeSort. Each merge operation has to put the biggest element into bottom of new stack
// so an extra reverse operation is needed. Time complexity is O(nlogn), space : O(nlgn)
import java.util.Stack;
import java.util.Random;

public class stackSort {
	public static<T extends Comparable<? super T>> Stack<T> insertSort(Stack<T> stack) {
		Stack<T> sorted = new Stack<T>();
		T temp;

chouclee

import java.util.Stack;

public class MyQueue<T> {
	private Stack<T> pushStack;
	private Stack<T> popStack;
	
	public MyQueue() {
		pushStack = new Stack<T>();
		popStack = new Stack<T>();
	}