Reverse a sublist of linked list

We are given a linked list and positions m and n. We need to reverse the linked list from position m to n.

Examples:

Input : 10->20->30->40->50->60->70->NULL
        m = 3, n = 6
Output : 10->20->60->50->40->30->70->NULL

Input :  1->2->3->4->5->6->NULL 
         m = 2, n = 4
Output : 1->4->3->2->5->6->NULL

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

To reverse the linked list from position m to n, we find addresses of start and end position of the linked list by running a loop, and then we unlink this part from the rest of the list and then use the normal linked list reverse function which we have earlier used for reversing the complete linked list, and use it to reverse the portion of the linked list which need to be reversed. After reversal, we again attach the portion reversed to the main list.

// C program to reverse a linked list 
// from position m to position n 
#include <stdio.h> 
#include <stdlib.h> 
  
// Linked list node 
struct Node { 
    int data; 
    struct Node* next; 
}; 
  
// the standard reverse function used 
// to reverse a linked list 
struct Node* reverse(struct Node* head) 
{ 
    struct Node* prev = NULL;     
    struct Node* curr = head; 
  
    while (curr) { 
        struct Node* next = curr->next; 
        curr->next = prev; 
        prev = curr; 
        curr = next; 
    } 
    return prev; 
} 
  
// function used to reverse a linked list 
// from position m to n which uses reverse 
// function 
Node* reverseBetween(Node* head, int m, int n) 
{ 
    if (m == n) 
        return head; 
  
    // revs and revend is start and end respectively 
    // of the portion of the linked list which 
    // need to be reversed. revs_prev is previous 
    // of starting position and revend_next is next 
    // of end of list to be reversed. 
    Node* revs = NULL, *revs_prev = NULL; 
    Node* revend = NULL, *revend_next = NULL; 
  
    // Find values of above pointers. 
    int i = 1; 
    Node* curr = head; 
    while (curr && i <= n) { 
        if (i < m) 
            revs_prev = curr; 
  
        if (i == m) 
            revs = curr; 
  
        if (i == n) { 
            revend = curr; 
            revend_next = curr->next; 
        } 
  
        curr = curr->next; 
        i++; 
    } 
    revend->next = NULL; 
  
    // Reverse linked list starting with 
    // revs. 
    revend = reverse(revs); 
  
    // If starting position was not head 
    if (revs_prev) 
        revs_prev->next = revend; 
  
    // If starting position was head 
    else
        head = revend; 
  
    revs->next = revend_next; 
    return head; 
} 
  
void print(struct Node* head) 
{ 
    while (head != NULL) { 
        printf("%d ", head->data); 
        head = head->next; 
    } 
    printf("\n"); 
} 
  
// function to add a new node at the 
// beginning of the list 
void push(struct Node** head_ref, int new_data) 
{ 
    struct Node* new_node = new Node; 
    new_node->data = new_data; 
    new_node->next = (*head_ref); 
    (*head_ref) = new_node; 
} 
  
// Driver code 
int main() 
{ 
    struct Node* head = NULL; 
    push(&head, 70); 
    push(&head, 60); 
    push(&head, 50); 
    push(&head, 40); 
    push(&head, 30); 
    push(&head, 20); 
    push(&head, 10); 
    reverseBetween(head, 3, 6); 
    print(head); 
    return 0; 
} 

Output:

10 20 60 50 40 30 70

This article is contributed by Akshit Agarwal. 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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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