Given a singly linked list, we need to arrange the consonants and vowel nodes of it in such a way that all the vowels nodes come before the consonants while maintaining the **order of their arrival**.

Examples:

Input : a -> b -> c -> e -> d -> o -> x -> i Output : a -> e -> o -> i -> b -> c -> d -> x

**Solution :**

The idea is to keep a marker of the latest vowel we found while traversing the list. If we find another vowel, we take it out of the chain and put it after the existing latest vowel. Example: For linked list:

a -> b -> c -> e -> d -> o -> x -> i

say our latestVowel reference references the ‘a’ node, and that we currently reached the ‘e’ node. We do:

a -> e -> b -> c -> d -> o -> x -> i

So what was after the ‘a’ node is now after the ‘e’ node after deleting it, and linking ‘a’ directly to ‘e’.

To properly remove and add links, it’s best to use the node before the one you are checking. So if you have a *curr*, you will check *curr->next* node to see if it’s a vowel or not. If it is, we need to add it after the latestVowel node and then it’s easy to remove it from the chain by assigning its next to curr’s next. Also if a list only contains consonants, we simply return head.

/* C++ program to arrange consonants and vowels nodes in a linked list */ #include<bits/stdc++.h> using namespace std; /* A linked list node */ struct Node { char data; struct Node *next; }; /* Function to add new node to the List */ Node *newNode(char key) { Node *temp = new Node; temp->data = key; temp->next = NULL; return temp; } // utility function to print linked list void printlist(Node *head) { if (! head) { cout << "Empty List\n"; return; } while (head != NULL) { cout << head->data << " "; if (head->next) cout << "-> "; head = head->next; } cout << endl; } // utility function for checking vowel bool isVowel(char x) { return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u'); } /* function to arrange consonants and vowels nodes */ Node *arrange(Node *head) { Node *newHead = head; // for keep track of vowel Node *latestVowel; Node *curr = head; // list is empty if (head == NULL) return NULL; // We need to discover the first vowel // in the list. It is going to be the // returned head, and also the initial // latestVowel. if (isVowel(head->data)) // first element is a vowel. It will // also be the new head and the initial // latestVowel; latestVowel = head; else { // First element is not a vowel. Iterate // through the list until we find a vowel. // Note that curr points to the element // *before* the element with the vowel. while (curr->next != NULL && !isVowel(curr->next->data)) curr = curr->next; // This is an edge case where there are // only consonants in the list. if (curr->next == NULL) return head; // Set the initial latestVowel and the // new head to the vowel item that we found. // Relink the chain of consonants after // that vowel item: // old_head_consonant->consonant1->consonant2-> // vowel->rest_of_list becomes // vowel->old_head_consonant->consonant1-> // consonant2->rest_of_list latestVowel = newHead = curr->next; curr->next = curr->next->next; latestVowel->next = head; } // Now traverse the list. Curr is always the item // *before* the one we are checking, so that we // can use it to re-link. while (curr != NULL && curr->next != NULL) { if (isVowel(curr->next->data)) { // The next discovered item is a vowel if (curr == latestVowel) { // If it comes directly after the // previous vowel, we don't need to // move items around, just mark the // new latestVowel and advance curr. latestVowel = curr = curr->next; } else { // But if it comes after an intervening // chain of consonants, we need to chain // the newly discovered vowel right after // the old vowel. Curr is not changed as // after the re-linking it will have a // new next, that has not been checked yet, // and we always keep curr at one before // the next to check. Node *temp = latestVowel->next; // Chain in new vowel latestVowel->next = curr->next; // Advance latestVowel latestVowel = latestVowel->next; // Remove found vowel from previous place curr->next = curr->next->next; // Re-link chain of consonants after latestVowel latestVowel->next = temp; } } else { // No vowel in the next element, advance curr. curr = curr->next; } } return newHead; } // Driver code int main() { Node *head = newNode('a'); head->next = newNode('b'); head->next->next = newNode('c'); head->next->next->next = newNode('e'); head->next->next->next->next = newNode('d'); head->next->next->next->next->next = newNode('o'); head->next->next->next->next->next->next = newNode('x'); head->next->next->next->next->next->next->next = newNode('i'); printf("Linked list before :\n"); printlist(head); head = arrange(head); printf("Linked list after :\n"); printlist(head); return 0; }

Output:

Linked list before : a -> b -> c -> e -> d -> o -> x -> i Linked list after : a -> e -> o -> i -> b -> c -> d -> x

References:

Stackoverflow

This article is contributed by **Gaurav Miglani**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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