链表操作

链表的结点结构

 struct Node
 {
 int data ;
 Node *next ;
 };
 typedef struct Node Node ;

(1)已知链表的头结点head,写一个函数把这个链表逆序 ( Intel)

 Node * ReverseList(Node *head) //链表逆序
 {
 if ( head == NULL || head->next == NULL )
 return head;
 Node *p1 = head ;
 Node *p2 = p1->next ;
 Node *p3 = p2->next ;
 p1->next = NULL ;
 while ( p3 != NULL )
 {
 p2->next = p1 ;
 p1 = p2 ;
 p2 = p3 ;
 p3 = p3->next ;
 }
 p2->next = p1 ;
 head = p2 ;
 return head ;
 }

(2)已知两个链表head1 和head2 各自有序，请把它们合并成一个链表依然有序。(保留所有结点，即便大小相同）

 Node * Merge(Node *head1 , Node *head2)
 {
 if ( head1 == NULL)
 return head2 ;
 if ( head2 == NULL)
 return head1 ;
 Node *head = NULL ;
 Node *p1 = NULL;
 Node *p2 = NULL;
 if ( head1->data < head2->data )
 {
 head = head1 ;
 p1 = head1->next;
 p2 = head2 ;
 }
 else
 {
 head = head2 ;
 p2 = head2->next ;
 p1 = head1 ;
 }
 Node *pcurrent = head ;
 while ( p1 != NULL && p2 != NULL)
 {
 if ( p1->data <= p2->data )
 {
 pcurrent->next = p1 ;
 pcurrent = p1 ;
 p1 = p1->next ;
 }
 else
 {
 pcurrent->next = p2 ;
 pcurrent = p2 ;
 p2 = p2->next ;
 }
 }
 if ( p1 != NULL )
 pcurrent->next = p1 ;
 if ( p2 != NULL )
 pcurrent->next = p2 ;
 return head ;
 }

(3)已知两个链表head1 和head2 各自有序，请把它们合并成一个链表依然有序，这次要求用递归方法进行。 (Autodesk)

 Node * MergeRecursive(Node *head1 , Node *head2)
 {
 if ( head1 == NULL )
 return head2 ;
 if ( head2 == NULL)
 return head1 ;
 Node *head = NULL ;
 if ( head1->data < head2->data )
 {
 head = head1 ;
 head->next = MergeRecursive(head1->next,head2);
 }
 else
 {
 head = head2 ;
 head->next = MergeRecursive(head1,head2->next);
 }
 return head ;