Move Pointer Back to the Beginning of a Single Linked List

[gzeng]

Is there any simple way to move the next pointer back to the beginning of a single linked list after it travers thru the list?

[ravenspoint]

Set it equal to a pointer to the head of the list, which you have kept somewhere convenient.

[gzeng]

Set it equal to a pointer to the head of the list, which you have kept somewhere convenient.

Thanks for your instant reply.
How do I do it? In the following is the piece of code. At end I need to point head back to 10 not 6.

************************************************** ****************************

void main() {
   node * curr, * head, *reversed_node, *sorted_node, *temp1, *temp2;
   int i;
 
   head = NULL;
 
   //build up head: 10, 9, ..., 1
   for(i=1;i<=10;i++) {
      curr = (node *)malloc(sizeof(node));
      curr->val = i;
      curr->next  = head;
      head = curr;
   }
 
   // Delete a node with val 5
   temp1 = head;
 
   // At end, temp1 is the node preceding the one with val 5
   while (temp1->next != NULL & temp1->next->val != 5)
          temp1 = temp1->next;
 
   cout << temp1->val << endl;      // print out 6
 
   // At end head points to 6 not 10!
   while (head) {
       if (head == temp1) {
           head -> next = temp1->next->next;
           break;
       }
 
       head = head->next;
  }

[sicarie]

gzeng-

Please use code tags when you post (They're located on the right side of the page when you reply or start a thread). Thanks!

[gzeng]

I found the solution myself. At end, I added:

head = curr;

Thanks for your instant reply.
How do I do it? In the following is the piece of code. At end I need to point head back to 10 not 6.

************************************************** ****************************

void main() {
   node * curr, * head, *reversed_node, *sorted_node, *temp1, *temp2;
   int i;
 
   head = NULL;
 
   //build up head: 10, 9, ..., 1
   for(i=1;i<=10;i++) {
      curr = (node *)malloc(sizeof(node));
      curr->val = i;
      curr->next  = head;
      head = curr;
   }
 
   // Delete a node with val 5
   temp1 = head;
 
   // At end, temp1 is the node preceding the one with val 5
   while (temp1->next != NULL & temp1->next->val != 5)
          temp1 = temp1->next;
 
   cout << temp1->val << endl;      // print out 6
 
   // At end head points to 6 not 10!
   while (head) {
       if (head == temp1) {
           head -> next = temp1->next->next;
           break;
       }
 
       head = head->next;
  }