Develop Reference

Radix(Trie) Tree implementation for Cutomer search in Java

I am working on a project and need to search in data of millions of customers. I want to implement radix(trie) search algorithm. I have read and implement radix for a simple string collections. But Here I have a collection of customers and want to search it by name or by mobile number.
Customer Class:
public class Customer {
String name;
String mobileNumer;
public Customer (String name, String phoneNumer) {
this.name = name;
this.mobileNumer = phoneNumer;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getPhoneNumer() {
return mobileNumer;
}
public void setPhoneNumer(String phoneNumer) {
this.mobileNumer = phoneNumer;
}
}
RadixNode Class:
import java.util.HashMap;
import java.util.Map;
class RadixNode {
private final Map<Character, RadixNode> child = new HashMap<>();
private final Map<Customer, RadixNode> mobileNum = new HashMap<>();
private boolean endOfWord;
Map<Character, RadixNode> getChild() {
return child;
}
Map<Customer, RadixNode> getChildPhoneDir() {
return mobileNum;
}
boolean isEndOfWord() {
return endOfWord;
}
void setEndOfWord(boolean endOfWord) {
this.endOfWord = endOfWord;
}
}
Radix Class:
class Radix {
private RadixNode root;
Radix() {
root = new RadixNode();
}
void insert(String word) {
RadixNode current = root;
for (int i = 0; i < word.length(); i++) {
current = current.getChild().computeIfAbsent(word.charAt(i), c -> new RadixNode());
}
current.setEndOfWord(true);
}
void insert(Customer word) {
RadixNode current = root;
System.out.println("==========================================");
System.out.println(word.mobileNumer.length());
for (int i = 0; i < word.mobileNumer.length(); i++) {
current = current.getChildPhoneDir().computeIfAbsent(word.mobileNumer.charAt(i), c -> new RadixNode());
System.out.println(current);
}
current.setEndOfWord(true);
}
boolean delete(String word) {
return delete(root, word, 0);
}
boolean containsNode(String word) {
RadixNode current = root;
for (int i = 0; i < word.length(); i++) {
char ch = word.charAt(i);
RadixNode node = current.getChild().get(ch);
if (node == null) {
return false;
}
current = node;
}
return current.isEndOfWord();
}
boolean isEmpty() {
return root == null;
}
private boolean delete(RadixNode current, String word, int index) {
if (index == word.length()) {
if (!current.isEndOfWord()) {
return false;
}
current.setEndOfWord(false);
return current.getChild().isEmpty();
}
char ch = word.charAt(index);
RadixNode node = current.getChild().get(ch);
if (node == null) {
return false;
}
boolean shouldDeleteCurrentNode = delete(node, word, index + 1) && !node.isEndOfWord();
if (shouldDeleteCurrentNode) {
current.getChild().remove(ch);
return current.getChild().isEmpty();
}
return false;
}
public void displayContactsUtil(RadixNode curNode, String prefix)
{
// Check if the string 'prefix' ends at this Node
// If yes then display the string found so far
if (curNode.isEndOfWord())
System.out.println(prefix);
// Find all the adjacent Nodes to the current
// Node and then call the function recursively
// This is similar to performing DFS on a graph
for (char i = 'a'; i <= 'z'; i++)
{
RadixNode nextNode = curNode.getChild().get(i);
if (nextNode != null)
{
displayContactsUtil(nextNode, prefix + i);
}
}
}
public boolean displayContacts(String str)
{
RadixNode prevNode = root;
// 'flag' denotes whether the string entered
// so far is present in the Contact List
String prefix = "";
int len = str.length();
// Display the contact List for string formed
// after entering every character
int i;
for (i = 0; i < len; i++)
{
// 'str' stores the string entered so far
prefix += str.charAt(i);
// Get the last character entered
char lastChar = prefix.charAt(i);
// Find the Node corresponding to the last
// character of 'str' which is pointed by
// prevNode of the Trie
RadixNode curNode = prevNode.getChild().get(lastChar);
// If nothing found, then break the loop as
// no more prefixes are going to be present.
if (curNode == null)
{
System.out.println("No Results Found for \"" + prefix + "\"");
i++;
break;
}
// If present in trie then display all
// the contacts with given prefix.
System.out.println("Suggestions based on \"" + prefix + "\" are");
displayContactsUtil(curNode, prefix);
// Change prevNode for next prefix
prevNode = curNode;
}
for ( ; i < len; i++)
{
prefix += str.charAt(i);
System.out.println("No Results Found for \"" + prefix + "\"");
}
return true;
}
public void displayContactsUtil(RadixNode curNode, String prefix, boolean isPhoneNumber)
{
// Check if the string 'prefix' ends at this Node
// If yes then display the string found so far
if (curNode.isEndOfWord())
System.out.println(prefix);
// Find all the adjacent Nodes to the current
// Node and then call the function recursively
// This is similar to performing DFS on a graph
for (char i = '0'; i <= '9'; i++)
{
RadixNode nextNode = curNode.getChildPhoneDir().get(i);
if (nextNode != null)
{
displayContactsUtil(nextNode, prefix + i);
}
}
}
public boolean displayContacts(String str, boolean isPhoneNumber)
{
RadixNode prevNode = root;
// 'flag' denotes whether the string entered
// so far is present in the Contact List
String prefix = "";
int len = str.length();
// Display the contact List for string formed
// after entering every character
int i;
for (i = 0; i < len; i++)
{
// 'str' stores the string entered so far
prefix += str.charAt(i);
// Get the last character entered
char lastChar = prefix.charAt(i);
// Find the Node corresponding to the last
// character of 'str' which is pointed by
// prevNode of the Trie
RadixNode curNode = prevNode.getChildPhoneDir().get(lastChar);
// If nothing found, then break the loop as
// no more prefixes are going to be present.
if (curNode == null)
{
System.out.println("No Results Found for \"" + prefix + "\"");
i++;
break;
}
// If present in trie then display all
// the contacts with given prefix.
System.out.println("Suggestions based on \"" + prefix + "\" are");
displayContactsUtil(curNode, prefix, isPhoneNumber);
// Change prevNode for next prefix
prevNode = curNode;
}
for ( ; i < len; i++)
{
prefix += str.charAt(i);
System.out.println("No Results Found for \"" + prefix + "\"");
}
return true;
}
}
I have tried to search in a collection but got stuck. Any help / suggestion would be appreciated.

I propose you 2 ways of doing it.
First way: with a single trie.
It is possible to store all you need in a single trie. Your customer class is fine, and here is a possible RadixNode implementation.
I consider that there cannot be two customers with the same name, or with the same phone number. If it is not the case (possibility to have people with same name and different phone nb for instance) tell me in a comment I'll edit.
The thing that is important to understand, is that if you want to have two different ways of finding a customer, and you use a single trie, each customer will appear twice in your trie. Once at the end of the path corresponding to its name, and once after the end of the path corresponding to its phone number.
import java.util.HashMap;
import java.util.Map;
class RadixNode {
private Map<Character, RadixNode> children;
private Customer customer;
public RadixNode(){
this.children = new Map<Character, RadixNode>();
this.Customer = NULL;
}
Map<Character, RadixNode> getChildren() {
return children;
}
boolean hasCustomer() {
return this.customer != NULL;
}
Customer getCustomer() {
return customer;
}
void setCustomer(Customer customer) {
this.customer = customer;
}
}
As you can see, there is only one map storing the node's children. That is because we can see a phone number as a string of digits, so this trie will store all the customers ... twice. Once per name, once per phone number.
Now let's see an insert function. Your trie will need a root,n let's call it root.
public void insert(RadixNode root, Customer customer){
insert_with_name(root, customer, 0);
insert_with_phone_nb(root, customer, 0);
}
public void insert_with_name(RadixNode node, Customer customer, int idx){
if (idx == customer.getName().length()){
node.setCustomer(customer);
} else {
Character current_char = customer.getName().chatAt(idx);
if (! node.getChlidren().containsKey(current_char){
RadixNode new_child = new RadixNode();
node.getChildren().put(current_char, new_child);
}
insert_with_name(node.getChildren().get(current_char), customer, idx+1);
}
}
The insert_with_phone_nb() method is similar. This will work as long as people has unique names, unique phone numbers, and that someone's name cannot be someone's phone number.
As you can see, the method is recursive. I advice you to build your trie structure (and generally, everything based on tree structures) recursively, as it makes for simpler, and generallay cleaner code.
The search function is almost a copy-paste of the insert function:
public void search_by_name(RadixNode node, String name, int idx){
// returns NULL if there is no user going by that name
if (idx == name.length()){
return node.getCustomer();
} else {
Character current_char = name.chatAt(idx);
if (! node.getChlidren().containsKey(current_char){
return NULL;
} else {
return search_by_name(node.getChildren().get(current_char), name, idx+1);
}
}
}
Second way: with 2 tries
The principle is the same, all you have to do is reuse the code above, but keep two distinct root nodes, each of them will build a trie (one for names, one for phone numbers).
The only difference will be the insert function (as it will call insert_with_name and insert_with_phone_nb with 2 different roots), and the search function which will have to search in the right trie as well.
public void insert(RadixNode root_name_trie, RadixNode root_phone_trie, Customer customer){
insert_with_name(root_name_trie, customer, 0);
insert_with_phone_nb(root_phone_trie, customer, 0);
}
Edit: After comment precising there might be customers with the same name, here is an alternative implementation, to allow a RadixNode to contain references toward several Customer.
Replace the Customer customer attribute in RadixNode by, for example, a Vector<Customer>. The methods will have to be modified accordingly of course, and a search by name will then return to you a vector of customers (possibly empty), since this search can then lead to several results.
In your case, I'd go for a single trie, containing vectors of customers. So you can have both a search by name and phone (cast the number as a String), and a single data structure to maintain.

insert method of the Trie data structure

I am trying ti implement the insert method of the Patricia Trie data structure but I have the feeling I wrote to many code lines. Please can someone tell me where can I call the method insert(TrieNode nodeRoot, String s) rekursiv?
Code:
private void insert(TrieNode nodeRoot, String s) {
int len1 = nodeRoot.value.length();
int len2 = s.length();
int len = Math.min(len1, len2);
for (int index = 0; index < len; index++) {
if (s.charAt(index) != nodeRoot.value.charAt(index)) {
// In case the both words have common substrings and after the
// common substrings the words are split.
String samesubString = s.substring(0, index);
String substringSplit1 = nodeRoot.value.substring(index);
String substringSplit2 = s.substring(index);
if (!samesubString.isEmpty()) {
nodeRoot.value = samesubString;
}
TrieNode nodeLeft = new TrieNode(substringSplit1);
nodeLeft.isWord = true;
TrieNode nodeRight = new TrieNode(substringSplit2);
nodeRight.isWord = true;
if (nodeRoot.getNext() != null && !nodeRoot.getNext().isEmpty()) {
checkTheValieAvialable(nodeRoot, s, nodeRight);
} else {
nodeRoot.next.add(nodeLeft);
nodeRoot.next.add(nodeRight);
for (TrieNode subword : nodeRoot.getNext()) {
System.out.println(nodeRoot.getValue() + "---"
+ subword.getValue());
}
}
break;
} else if (index == (s.length() - 1)
|| index == (nodeRoot.value.length() - 1)) {
// In case the node just needs one path since one word is
// substring of the other.
// For example (aba and abac)
if (len1 > len2) {
// root value is longer
System.out.println("root value is longer");
String samesubString = nodeRoot.value.substring(0,
index + 1);
String different = nodeRoot.value.substring(index + 1);
if (nodeRoot.getNext() != null
&& !nodeRoot.getNext().isEmpty()) {
for (TrieNode subword : nodeRoot.getNext()) {
String subword2 = subword.getValue();
boolean contains = different.contains(subword2);
if (contains) {
String[] split = different.split(subword2);
TrieNode leaf1 = new TrieNode(split[1]);
leaf1.isWord = true;
subword.next.add(leaf1);
System.out.println("Test.");
}
}
} else {
String substringSplit1 = nodeRoot.value.substring(index + 1);
nodeRoot.value = samesubString;
TrieNode leaf = new TrieNode(substringSplit1);
leaf.isWord = true;
nodeRoot.next.add(leaf);
for (TrieNode subword : nodeRoot.getNext()) {
System.out.println(nodeRoot.getValue() + "---"
+ subword.getValue());
}
}
String substringSplit1 = nodeRoot.value
.substring(index + 1);
nodeRoot.value = samesubString;
nodeRoot.isWord = true;
TrieNode leaf = new TrieNode(substringSplit1);
leaf.isWord = true;
nodeRoot.next.add(leaf);
for (TrieNode subword : nodeRoot.getNext()) {
System.out.println(nodeRoot.getValue() + "---"
+ subword.getValue());
}
} else {
// new inserted string value is longer. For example (abac and aba).
System.out.println("instered is longer");
String samesubString = s.substring(0, index + 1);
String different = s.substring(index + 1);
if (nodeRoot.getNext() != null
&& !nodeRoot.getNext().isEmpty()) {
for (TrieNode subword : nodeRoot.getNext()) {
String subword2 = subword.getValue();
boolean contains = different.contains(subword2);
if (contains) {
String[] split = different.split(subword2);
TrieNode leaf1 = new TrieNode(split[1]);
leaf1.isWord = true;
subword.next.add(leaf1);
System.out.println("Test.");
}
}
} else {
String substringSplit1 = s.substring(index + 1);
s = samesubString;
TrieNode parentLeaf = new TrieNode(s);
parentLeaf.isWord = true;
TrieNode leaf = new TrieNode(substringSplit1);
leaf.isWord = true;
nodeRoot.next.add(leaf);
for (TrieNode subword : nodeRoot.getNext()) {
System.out.println(nodeRoot.getValue() + "---"
+ subword.getValue());
}
}
}
} else {
System.out.println("They are the same - " + index);
}
}
}
TrieNode class:
package patriciaTrie;
import java.util.ArrayList;
public class TrieNode {
ArrayList<TrieNode> next = new ArrayList<TrieNode>();
String value;
boolean isWord;
TrieNode(String value){
this.value = value;
}
public ArrayList<TrieNode> getNext() {
return next;
}
public void setNext(ArrayList<TrieNode> next) {
this.next = next;
}
public String getValue() {
return value;
}
public void setValue(String value) {
this.value = value;
}
}

While using recursion please consider the steps:
Base condition
Logic (if any)
Recursive call.
Ex. for factorial of number:
int fact(int n)
{
if(n==0 || n==1)
return 1; // Base condition
return n * fact(n-1); // Recursive call
}
Applying the same concept in Trie:
base condition is: while traversing through a path, if we have reached leaf, current string is not in trie, then create a new edge or node and add remaining character to it.
Recursively call the insert if we have found a matching node. And if a matching node doen't exist create a new path with common parent.
You can take help from link : http://www.geeksforgeeks.org/trie-insert-and-search/
The best way to approach to problem recursively is to identify base condition in a problem.

Java linked list - issues w 3 methods

I am currently working on a linked list project but I am stumped on my last 3 methods, my removeWord(), concatenate() and doubleChar().. I was wondering if anyone one could give me some tips in what I am doing wrong.
class Node {
private char letter;
private Node next;
public Node(char ch, Node link) {
letter = ch;
next = link;
}
public void setLetter(char letter) {
this.letter=letter;
}
public char getLetter() {
return letter;
}
public void setNext(Node next) {
this.next=next;
}
public Node getNext() {
return next;
}
}
class Word {
// instance variable pointing to the head of the linked list
private Node head;
// default constructor
public Word() {
head = null;
}
// copy constructor
public Word(Word w) {
this.head = copy(w.head);
}
private static Node copy(Node l){
Node newL;
if( l ==null){
newL=null;
}else{
newL = new Node(l.getLetter(),copy(l.getNext()));
}
return newL;
}
// constructor from a String
public Word( String s ) {
Node pt;
head = null;
for( int i = s.length()-1; i >=0; i--){
pt = new Node(s.charAt(i),head);
head = pt;
}
}
// for output purposes -- override Object version
// no spaces between the characters, no linefeeds/returns
#Override
public String toString() {
//s.charAt
return toString(head);
// return toString(head);
}
private static String toString(Node L){
String Word="";
if (L == null){
// do nothing
}
else{
Word = L.getLetter() + toString(L.getNext());
// return the letter
}
return Word;
}
// remove the first occurrence of the Word w from this Word
public void removeWord( Word w ) {
head = removeWord(head,w);
}
private static Node removeWord(Node L, Word w)
{
if(L == null)
{
// do nothing
}else if(L==w.head){
L = L.getNext();
}
else {
// remove the word
L.setNext(removeWord(L.getNext(),w));
}
return L;
}
// concatenate a copy of s to the end of this Word
public void concatenate( Word s ) {
this.head = concatenate(head,s);
}
private static Node concatenate(Node L, Word s){
if( L==null){
L = null;
}
else{
L = new Node(L.getLetter(),concatenate(L.getNext(),s));
L.setNext(concatenate(L.getNext(),s));// add to the end of the list
}
return L;
}
// make a copy of every occurrence of ch in this Word
// for example, if this Word is abbcbccb, doubleChar ( 'b' ) should
// change the Word to abbbbcbbccbb
public void doubleChar( char ch ) {
head = doubleChar(head,ch);
}
public static Node doubleChar(Node L, char ch){
if(L == null)
{
}
else if(L.getLetter()==ch){
}
else{
//double
L.setNext(doubleChar(L.getNext(),ch));
}
return L;
}

Java Binary Tree entered in a specific order

I am trying to complete an assignment where I need to write a Java program to take a string from the command line, and implement it as a Binary Tree in a specific order, then get the depth of the binary tree.
For example: "((3(4))7((5)9))"
would be entered as a tree with 7 as the root, 3 and 9 as the children, and 4 as a right child of 3, and 5 as a left child of 9.
My code is below.. The problem I am having is that, because I am basing my checks off of finding a right bracket, I am unsure how to get the elements correctly when they are not directly preceding the brackets, such as the 3 in the above string. Any direction would be greatly appreciated..
class Node {
int value;
Node left, right;
}
class BST {
public Node root;
// Add Node to Tree
public void add(int n) {
if (root == null) {
root = new Node( );
root.value = n;
}
else {
Node marker = root;
while (true) {
if (n < marker.value) {
if (marker.left == null) {
marker.left = new Node( );
marker.left.value = n;
break;
} else {
marker = marker.left;
}
} else {
if (marker.right == null) {
marker.right = new Node( );
marker.right.value = n;
break;
} else {
marker = marker.right;
}
}
}
}
} // End ADD
//Find Height of Tree
public int height(Node t) {
if (t.left == null && t.right == null) return 0;
if (t.left == null) return 1 + height(t.right);
if (t.right == null) return 1 + height(t.left);
return 1 + Math.max(height(t.left), height(t.right));
} // End HEIGHT
// Check if string contains an integer
public static boolean isInt(String s) {
try {
Integer.parseInt(s);
}
catch(NumberFormatException e) {
return false;
}
return true;
} // End ISINT
public int elementCount(String[] a) {
int count = 0;
for (int i = 0; i < a.length; i++) {
if (isInt(a[i])) count++;
}
return count;
}
} // End BST Class
public class Depth {
public static void main(String[] args) {
String[] a = args[0].split(" ");
BST tree = new BST();
int[] bcount = new int[10];
int[] elements = new int[10];
int x = 0, bracketcount = 0;
// Display entered string
System.out.print("Entered Format: ");
for (int j=0; j < a.length; j++) {
System.out.print(a[j]);
}
for (int i=0; i < a.length; i++) {
char c = a[i].charAt(0);
switch (c)
{
case '(':
bracketcount++;
break;
case ')':
if (isInt(a[i-1])) {
bcount[x] = bracketcount--;
elements[x++] = Integer.parseInt(a[i-1]);
}
break;
case '1':
case '7':
default : // Illegal character
if ( (a[i-1].charAt(0) == ')') && (a[i+1].charAt(0) == '(') ) {
bcount[x] = bracketcount;
elements[x++] = Integer.parseInt(a[i]);
}
break;
}
}
System.out.println("\nTotal elements: " + tree.elementCount(a));
// Display BracketCounts
for (int w = 0; w < x; w++) {
System.out.print(bcount[w] + " ");
}
System.out.println(" ");
// Display Elements Array
for (int w = 0; w < x; w++) {
System.out.print(elements[w] + " ");
}
System.out.println("\nDepth: " + tree.height(tree.root));
// Build the tree
for (int y = 0; y < x-1; y++) {
for (int z = 1; z < tree.height(tree.root); z++) {
if (bcount[y] == z) {
tree.add(elements[y]);
}
}
}
} // End Main Function
public static boolean isInt(String s) {
try {
Integer.parseInt(s);
}
catch(NumberFormatException e) {
return false;
}
return true;
}
} // End Depth Class

I would do a couple of statements to get access to a tree with that kind of shape:
For input string : input= "((3(4))7((5)9))"
You could do :
public class Trial {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String input = "((3(4))7((5)9))";
String easier = input.replaceAll("\\(\\(", "");
String evenEasier = easier.replaceAll("\\)\\)", "");
System.out.println(evenEasier);
int firstVal = Integer.parseInt(evenEasier.substring(0, 1));
int firstBracketVal = Integer.parseInt(evenEasier.substring(2, 3));
int middleVal = Integer.parseInt(evenEasier.substring(3, 4));
int secondBracketVal = Integer.parseInt(evenEasier.substring(4,5));
int lastVal = Integer.parseInt(evenEasier.substring(6));
System.out.println("First Val:"+firstVal);
System.out.println("First bracket Val:"+firstBracketVal);
System.out.println("Middle Val:"+middleVal);
System.out.println("Second Bracket Val:"+secondBracketVal);
System.out.println("Last Val:"+lastVal);
}
}
This however would only ever work for entries in that specific format, if that were to change, or the length of the input goes up - this would work a bit or break.....If you need to be able to handle more complicated trees as input in this format a bit more thought would be needed on how to best handle and convert into your internal format for processing.

pseudocode:
function getNode(Node)
get one char;
if (the char is "(")
getNode(Node.left);
get one char;
end if;
Node.value = Integer(the char);
get one char;
if (the char is "(")
getNode(Node.right);
get one char;
end if;
//Now the char is ")" and useless.
end function
Before calling this function, you should get a "(" first.
In this method, the framwork of a Node in string is "[leftchild or NULL] value [rightchild or NULL])".
"("is not belong to the Node, but ")" is.

Assignment on sorting a list

I'm very close to being done, but can't quite figure out how to tie everything together. I have the separate methods responsible for their particular task, but i feel like my cohesion is really bad. not real sure how they tie together and what needs to be called in main. The goal here is to read a text file from the command line and list the words in the story lexicographically.
% java Ten < story.txt
Word Occurs
==== ======
a 21
animal 3
.
.
.
zebra 1
%
Here's my code thus far:
import java.util.Scanner;
public class Ten2
{
public static void main(String [] arg)
{
Scanner input = new Scanner(System.in);
String word;
List sortedList = new List();
word = nextWord(input);
while (word!=null) {
sortedList.insert(word);
word = nextWord(input);
}
sortedList.printWords();
}
private static String nextWord(Scanner input)
{
// return null if there are no more words
if (input.hasNext() == false )
return null;
// take next element, convert to lowercase, store in s
else {
String s = input.next().toLowerCase() ;
// empty string
String token = "";
// loop through s and concatonate chars onto token
for (int i =0; i < s.length(); i++) {
if (Character.isLetter(s.charAt(i)) == true)
token = token + s.charAt(i);
else if (s.charAt(i) == '\'' )
token = token + s.charAt(i);
else if (s.charAt(i) == '-')
token = token + s.charAt(i);
}
return token;
}
}
}
class List
{
/*
* Internally, the list of strings is represented by a linked chain
* of nodes belonging to the class ListNode. The strings are stored
* in lexicographical order.
*/
private static class ListNode
{
// instance variables for ListNode objects
public String word;
public ListNode next;
public int count;
// Listnode constructor
public ListNode(String w, ListNode nxt)
{
word = w; // token from nextWord()?
next = nxt; // link to next ListNode
count = 1; // number of word occurences
}
}
// instance variables for List object
private ListNode first;
private int numWords;
// constructor postcondition: creates new Listnode storing object
public List()
{
first = null; // pointer to ListNode?
numWords = 0; // counter for nodes in list
}
// Insert a specified word into the list, keeping the list
// in lexicographical order.
public void insert(String word)
{
// check if first is null
if (first == null) {
ListNode newNode;
newNode = addNode(word, null);
first = newNode;
}
// else if (first is not null) check if word matches first word in List
else if (word.equals(first.word)) {
// increase count
first.count++;
}
// else (first is not null && doesn't match first word)
else {
ListNode newNode;
ListNode current;
current = first;
ListNode previous;
previous = null;
int cmp = word.compareTo(current.word);
/*
* Fist two cases of empty list and word already existing
* handled in above if and else statements, now by using compareTo()
* method between the words, the insertion postion can be determined.
* Links between ListNode variables current and previous need to be
* modified in order to maintain the list
*/
// loop as long as value comparing to is positive
// when compareTo() returns positive this means the "word" parameter is greater than the word in the list
while ((cmp >0) && (current.next != null)) {
previous = current;
current = current.next;
cmp = word.compareTo(current.word);
}
// insert after current at end of list
if ((cmp >0 && current.next == null)) {
newNode = addNode(word, null);
current.next = newNode;
}
// increments count when word already exists
else if (cmp==0) {
current.count++;
}
// else (cmp < 0) we insert BEFORE current
else {
newNode = addNode(word, current);
// first node in list comes after new word
if (previous == null) {
first = newNode;
}
else {
// inserting new word in middle of list
previous.next = newNode;
}
}
}
}
// method to add new ListNode and increase counter
private ListNode addNode(String word, ListNode next)
{
ListNode newNode = new ListNode(word, next);
numWords++;
return newNode;
}
// Returns a string array that contains all the words in the list.
public String[] getWords()
{
String[] Words = new String[numWords];
ListNode current = first;
int i =0;
while (current != null) {
Words[i] = current.word;
current = current.next;
i++;
}
return Words;
}
// Returns an int array that contains the number of times
// each word occurs in the list.
public int[] getNumbers()
{
int[] Numbers = new int[numWords];
ListNode current = first;
int i =0;
while (current != null) {
Numbers[i] = current.count;
current = current.next;
i++;
}
return Numbers;
}
// Outputs the string array and int array containing all the
// words in the list and the number of times each occurs.
public void printWords()
{
int[] Numbers = getNumbers();
String[] Words = getWords();
System.out.println("Word \t \t Occurs");
System.out.println("==== \t \t ======");
for (int i =0; i < numWords; i++) {
System.out.println(Words[i] + " \t " + Numbers[i]);
}
}
}

Well, I would start by defining what you want your program to do, which you've already done:
The goal here is to read a text file from the command line and list the words in the story lexicographically.
You're main function does this almost. Basically, what you need is a loop to tie it together:
public static void main(String [] arg)
{
// print out your initial information first (i.e. your column headers)
// ...
List sortedList = new List();
String word = nextWord();
// now, the question is... what is the end condition of the loop?
// probably that there aren't any more words, so word in this case
// will be null
while (word != null)
{
sortedList.insert(word);
word = nextWord();
}
// now, your list takes care of the sorting, so let's just print the list
sortedList.printWords();
}
I think that's all there is to it. Normally, I don't like to post solutions to homework questions, but in this case, since you already had all of the code and you just needed a little nudge to drive you in the right direction, I think it's fine.
There are a few things I noticed that are incorrect with your
Your list constructor has a 'void' return type - there should be no return type on constructors:
public List() //make an empty list
{
first = null;
numWords = 0;
}
The 'else' statement in this method is unneeded:
public static String nextWord()
{
if ( keyboard.hasNext() == false )
return null;
else {
String start = keyboard.next().toLowerCase() ;
String organized = "";
for (int i =0; i < start.length(); i++) {
if (Character.isLetter(start.charAt(i)) == true)
organized = organized + start.charAt(i);
else if (start.charAt(i) == '\'' )
organized = organized + start.charAt(i);
else if (start.charAt(i) == '-')
organized = organized + start.charAt(i);
}
return organized;
}
}
So, this should be:
public static String nextWord()
{
if ( keyboard.hasNext() == false )
return null;
String start = keyboard.next().toLowerCase() ;
String organized = "";
for (int i =0; i < start.length(); i++) {
if (Character.isLetter(start.charAt(i)) == true)
organized = organized + start.charAt(i);
else if (start.charAt(i) == '\'' )
organized = organized + start.charAt(i);
else if (start.charAt(i) == '-')
organized = organized + start.charAt(i);
}
return organized;
}
If you want to use a BufferedReader, it's pretty easy. Just set it up in your main method:
if (arg.length > 0)
{
// open our file and read everything into a string buffer
BufferedReader bRead = null;
try {
bRead = new BufferedReader(new FileReader(arg[0]));
} catch(FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
System.exit(0);
}
setupScanner(bRead);
}
Then, create a new method that sets up the scanner object:
public static void setupScanner(BufferedReader rdr)
{
keyboard = new Scanner(rdr);
}
And then just pass it in on the command line (i.e. java ten2 [filename])

import java.util.Scanner;
public class Ten2
{
public static void main(String [] arg)
{
Scanner input = new Scanner(System.in);
String word;
List sortedList = new List();
word = nextWord(input);
while (word!=null) {
sortedList.insert(word);
word = nextWord(input);
}
sortedList.printWords();
}
private static String nextWord(Scanner input)
{
// return null if there are no more words
if (input.hasNext() == false )
return null;
// take next element, convert to lowercase, store in s
else {
String s = input.next().toLowerCase() ;
// empty string
String token = "";
// loop through s and concatonate chars onto token
for (int i =0; i < s.length(); i++) {
if (Character.isLetter(s.charAt(i)) == true)
token = token + s.charAt(i);
else if (s.charAt(i) == '\'' )
token = token + s.charAt(i);
else if (s.charAt(i) == '-')
token = token + s.charAt(i);
}
return token;
}
}
}
class List
{
/*
* Internally, the list of strings is represented by a linked chain
* of nodes belonging to the class ListNode. The strings are stored
* in lexicographical order.
*/
private static class ListNode
{
// instance variables for ListNode objects
public String word;
public ListNode next;
public int count;
// Listnode constructor
public ListNode(String w, ListNode nxt)
{
word = w; // token from nextWord()?
next = nxt; // link to next ListNode
count = 1; // number of word occurences
}
}
// instance variables for List object
private ListNode first;
private int numWords;
// constructor postcondition: creates new Listnode storing object
public List()
{
first = null; // pointer to ListNode?
numWords = 0; // counter for nodes in list
}
// Insert a specified word into the list, keeping the list
// in lexicographical order.
public void insert(String word)
{
// check if first is null
if (first == null) {
ListNode newNode;
newNode = addNode(word, null);
first = newNode;
}
// else if (first is not null) check if word matches first word in List
else if (word.equals(first.word)) {
// increase count
first.count++;
}
// else (first is not null && doesn't match first word)
else {
ListNode newNode;
ListNode current;
current = first;
ListNode previous;
previous = null;
int cmp = word.compareTo(current.word);
/*
* Fist two cases of empty list and word already existing
* handled in above if and else statements, now by using compareTo()
* method between the words, the insertion postion can be determined.
* Links between ListNode variables current and previous need to be
* modified in order to maintain the list
*/
// loop as long as value comparing to is positive
// when compareTo() returns positive this means the "word" parameter is greater than the word in the list
while ((cmp >0) && (current.next != null)) {
previous = current;
current = current.next;
cmp = word.compareTo(current.word);
}
// insert after current at end of list
if ((cmp >0 && current.next == null)) {
newNode = addNode(word, null);
current.next = newNode;
}
// increments count when word already exists
else if (cmp==0) {
current.count++;
}
// else (cmp < 0) we insert BEFORE current
else {
newNode = addNode(word, current);
// first node in list comes after new word
if (previous == null) {
first = newNode;
}
else {
// inserting new word in middle of list
previous.next = newNode;
}
}
}
}
// method to add new ListNode and increase counter
private ListNode addNode(String word, ListNode next)
{
ListNode newNode = new ListNode(word, next);
numWords++;
return newNode;
}
// Returns a string array that contains all the words in the list.
public String[] getWords()
{
String[] Words = new String[numWords];
ListNode current = first;
int i =0;
while (current != null) {
Words[i] = current.word;
current = current.next;
i++;
}
return Words;
}
// Returns an int array that contains the number of times
// each word occurs in the list.
public int[] getNumbers()
{
int[] Numbers = new int[numWords];
ListNode current = first;
int i =0;
while (current != null) {
Numbers[i] = current.count;
current = current.next;
i++;
}
return Numbers;
}
// Outputs the string array and int array containing all the
// words in the list and the number of times each occurs.
public void printWords()
{
int[] Numbers = getNumbers();
String[] Words = getWords();
System.out.println("Word \t \t Occurs");
System.out.println("==== \t \t ======");
for (int i =0; i < numWords; i++) {
System.out.println(Words[i] + " \t " + Numbers[i]);
}
}
}