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.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode *newnode1 = new ListNode(-1);
        ListNode *newnode2 = new ListNode(-1);
        ListNode  *smallr = newnode1;
        ListNode  *large = newnode2;
        ListNode *cur = head;
        
        
        while(cur !=NULL){
            if(cur->val
  
   next= cur;
                smallr=smallr->next;
                 
            }else{
                large->next =cur;
                large=large->next;
            }
            cur=cur->next;
        }
        large->next=NULL;  //处理好尾部节点  防止有环
        smallr->next= newnode2->next;
        return newnode1->next;
    }
};