Question

How would I make a linked list of a linked list in C? For example I want a linked list that holds all the words inputted from the user. Example:

typedef struct node{

char *word;

struct node *next;

}node;

This list will hold all the strings that were inputted from the user.

What I want to do is have another linked list that holds all the words that are no vowel matches which means that they would be equal if we remove the vowels (non-case sensitive) in some sort of set. Like if the user inputs:

hat hit hoot hate test tast past pest

The output would be:

hat hit hoot HATE

test tast

past pest

Every line of words are equal if we remove the vowels from them. All of which are stored in a linked list of their own.

I know this is somewhat easy to do in other languages that have object oriented, but C is not, and I am barely starting to learn it so any help would be appreciated!

Answer 1

You can use another struct on top of the node struct and create a "list" linked list.

typedef struct node{
    char *word;
    struct node *next;
}node;

typedef struct list{
    node* n;
    int alphabet[26];
    node* next;
}list;

I have not provided the code for the entire problem because I feel providing you with the clear logic of the process should be sufficient for you to program the code.

The logic for the separation of lists of strings based on your criteria is as follows

1. The List struct has a field called alphabet[26], this acts as a dictionary of characters, if the number of characters present in a single word (minus the vowels a,e,i,o,u) are the same then the string will be equal to its peers.
2. This alphabet array allows us to acheive exactly that. The array is essentially storing the frequencies of each of the 26 characters. The index 0 stores the frequency of a, 1 stores the frequency of b and so on.
3. Parse through the string input provided by the user. Assume the string input is " hat hit hoot hate test tast past pest ", as you had mentioned.
4. For each word (for example - hate), do the following
   1. Parse the word character by character and record the number of times a character appears (this will be akin to the alphabet array) and save this information in an array (let's call it temp)
   2. Now, use a while loop to go through the "list" linked list one by one and for e ach iteration check whether the alphabet array in the linked list item matched the temp array
   3. If the array matches (which can be done by checking each of the 21 elements, remember not checking for indices that correspond to vowels) then append this word to the "node" linked list inside this "list" linked list.
   4. For "hate" the array will be empty, except for indices 7 (corresponding to character h's frequency) and 19 (corresponding to character t's frequency)
   5. The above step may look like

      // listIterator was initialised as 
      // list* listIterator = (list*)malloc(sizeof(list))
      // this acts as the general way to traverse the linked list called "list"
      listIterator->next->next = (node*)malloc(sizeof(node))
      listIterator->next->next->word = //the word you are parsing through

   6. If the array does not match any of the linked list's alphabet arrays, then we have to create a new linked list node and then store this word by creating another linked list inside this new one (for storing list of words, that is)

The process may seem a bit confusing but the ideas are very simple, you have a list inside a list, both of which are mutable and extendable.