Creating an n array with a linked list of ints - c++

Question

I recently made an 26array and tried to simulate a dictionary.
I can't seem to figure out how to make this. I've tried to work with passing in a linkedlist of ints instead of a string. My current code creates 26 nodes(a-z) and then each of those nodes has 26 nodes(a-z). I would like to implement a way to do this with ints, say (1-26). These int nodes will represent items, and the linkedlist of ints I want to pass in will contain a set of ints that I want represented in the tree similar to a string.
Example: pass in the set {1, 6 , 8}, instead of a string such as "hello"
#include <iostream>
using namespace std;
class N26
{
private:
struct N26Node
{
bool isEnd;
struct N26Node *children[26];
}*head;
public:
N26();
~N26();
void insert(string word);
bool isExists(string word);
void printPath(char searchKey);
};
N26::N26()
{
head = new N26Node();
head->isEnd = false;
}
N26::~N26()
{
}
void N26::insert(string word)
{
N26Node *current = head;
for(int i = 0; i < word.length(); i++)
{
int letter = (int)word[i] - (int)'a';
if(current->children[letter] == NULL)
{
current->children[letter] = new N26Node();
}
current = current->children[letter];
}
current->isEnd = true;
}
/* Pre: A search key
* Post: True is the search key is found in the tree, otherwise false
* Purpose: To determine if a give data exists in the tree or not
******************************************************************************/
bool N26::isExists(string word)
{
N26Node *current = head;
for(int i=0; i<word.length(); i++)
{
if(current->children[((int)word[i]-(int)'a')] == NULL)
{
return false;
}
current = current->children[((int)word[i]-(int)'a')];
}
return current->isEnd;
}

Answer 1

class N26
{
private:
N26Node newNode(void);
N26Node *mRootNode;
...
};
N26Node *newNode(void)
{
N26Node *mRootNode = new N26Node;
mRootNode = NULL;
mRootNode->mData = NULL;
for ( int i = 0; i < 26; i++ )
mRootNode->mAlphabet[i] = NULL;
return mRootNode;
}
Ah! My eyes!
Seriously, you are attempting something much too advanced. Your code is full of bugs and cannot work as intended. Tinkering will not help, you must go back to basics of pointers and linked lists. Study the basics and do not attempt anything like a linked list of linked lists until you understand what is wrong with the code above.
I'll give you some hints: "memory leak", "dangling pointer", "type mismatch", "undefined behavior".

Answer 2

I didnt quite use linked lists, but I managed to get it working using arrays.
/* *** Author: Jamie Roland
* Class: CSI 281
* Institute: Champlain College
* Last Update: October 31, 2012
*
* Description:
* This class is to implement an n26 trie. The
* operations
* available for this impementation are:
*
* 1. insert
* 2. isEmpty
* 3. isExists
* 4. remove
* 5. showInOrder
* 6. showPreOrder
* 7. showPostOrder
*
* Certification of Authenticity:
* I certify that this assignment is entirely my own work.
**********************************************************************/
#include <iostream>
using namespace std;
class N26
{
private:
struct N26Node
{
bool isEnd;
struct N26Node *children[26];
}*head;
public:
N26();
~N26();
void insert(int word[]);
bool isExists(int word[]);
void printPath(char searchKey);
};
N26::N26()
{
head = new N26Node();
head->isEnd = false;
}
N26::~N26()
{
}
void N26::insert(int word[])
{
int size = sizeof word/sizeof(int);
N26Node *current = head;
for(int i = 0; i < size; i++)
{
int letter = word[i] - 1;
if(current->children[letter] == NULL)
{
current->children[letter] = new N26Node();
}
current = current->children[letter];
}
current->isEnd = true;
}
/* Pre: A search key
* Post: True is the search key is found in the tree, otherwise false
* Purpose: To determine if a give data exists in the tree or not
******************************************************************************/
bool N26::isExists(int word[])
{
int size = sizeof word/sizeof(int);
N26Node *current = head;
for(int i=0; i<size; i++)
{
if(current->children[(word[i]-1)] == NULL)
{
return false;
}
current = current->children[(word[i]-1)];
}
return current->isEnd;
}