The algorithm summarizes the single linked list in the way of the right and left half regions

Take the right side of the list and insert it into the linked list on the left.

Ideas:

If the list length is N, the method that gives the time complexity O (n) extra space complexity to O (1) is given directly

1 if the list is empty, or the length is 1 does not adjust

2 Linked list length is greater than 1 o'clock, traverse one to find the last node in the left half is recorded as mid

Package Tt;public class Test115 {public    class node{public    int value;    Public Node next;        public Node (int data) {    this.value=data;    }        }   public void Relocate (Node head) {     if (Head==null | | head.next==null) {      return;      }   Node mid = head;   Node right = Head.next;   while (Right.next!=null && right.next.next!=null) {   mid=mid.next;   Right=right.next.next;   }   Right=mid.next;   Mid.next=null;   MERGELR (head, right);   }      public void MERGELR (node-left, node-right) {   node-next = null;   while (left.next!=null) {   next=right.next;   Right.next=left.next;   Left.next=right;   Left=right.next;   Right=next;   }   left.next=right;   }       }

　　

The algorithm summarizes the single linked list in the way of the right and left half regions