*Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

public class Solution {
    public ListNode partition(ListNode head, int x) 
    {
        if (head == null) 
        {
            return null;
        }
        
        ListNode dummyLeft = new ListNode(0);
        ListNode dummyRight = new ListNode(0);
        ListNode Left = dummyLeft;
        ListNode Right = dummyRight;
        
        
        while(head!=null)
        {
            if(head.val<x)
            {
                Left.next = head;
                Left = head;  // 把头移到当前
                //Left = Left.next;
            }
            else 
            {
                Right.next = head;
                Right = head;  //
                //Right = Right.next;
            }
            head = head.next;
            
        }
        
        Right.next=null;
        Left.next = dummyRight.next;
        return dummyLeft.next;
        
    }

}