The algorithm summarizes the nodes that delete the middle node of the list and A/b node (the important idea of the middle node of the linked list)

Table header node for given list head, implementing function to delete the middle node of the linked list

Promotion: The head node of a given list, integer A and integer b, implements a function to delete a/b node

First to analyze the original problem,

Length 1 Direct return

Length 2 Remove the head node

Length 3 Remove the second length 4 remove the second length 5 remove the third one ... Length each additional 2 deleted nodes move back one node

If you want to delete a node, you need to find the previous node of the node to be deleted

Package Tt;public class Test87 {public class Node{public int value;public node next;public node (int data) {this.value = dat A;}} public static node Removemidnode (node head) {if (Head==null | | head.next = = NULL) {return head;} if (Head.next.next = = null) {return head.next;} Node pre = head; Node cur = head.next.next;while (cur.next!=null & Cur.next.next! = null) {Pre=pre.next;cur=cur.next.next;} Pre.next = Pre.next.next;return Head;}}

　　

The formula is calculated as Double R = ((double) (a*n))/((double) b) n is the length of the linked list is rounded up and then the first node of the node is found to be deleted.

 PackageTT;ImportTT. Test85.node; Public classTest87 { PublicNode Removebyratio (node head,intAintb) {                if(A<1 |a >b) {            returnHead//do not delete any nodes        }        intn = 0; Node cur=Head;  while(cur! =NULL) {n++;//CountCur =Cur.next; } N= (int) Math.ceil ((Double) (A*n))/(Double) b); if(n==1) {Head=Head.next; }        if(n>1) {cur=Head;  while(--n! = 1) {cur=Cur.next; } Cur.next=Cur.next.next; }        returnHead; }        }

The algorithm summarizes the nodes that delete the middle node of the list and A/b node (the important idea of the middle node of the linked list)