找出链表倒数第n个节点元素的二个方法

node* find_nth_to_last(node* head,int n) { if(head==NULL || n<1) return NULL; node*p,*q; p=q=head; while(q!=NULL && n--){ q=q->next; } if(n>=0) return NULL; while(p!=NULL && q!=NULL){ p=p->next; q=q->next; } return p; }

LinkedListNode nthToLast(LinkedListNode head, int n) { if (head == null || n < 1) { return null; } LinkedListNode p1 = head; LinkedListNode p2 = head; for (int j = 0; j < n - 1; ++j) { // skip n-1 steps ahead if (p2 == null) { return null; // not found since list size < n } p2 = p2.next; } while (p2.next != null) { p1 = p1.next; p2 = p2.next; } return p1; }