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.

/* *
 * 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) 
    {
         if(head==NULL || head->next==NULL)  return head;
        
        ListNode* pLarge;
        ListNode* pSmall;
        ListNode* pMid;
        
         if(head->val<x) pSmall=head;
         else
        {
            pLarge=head;
            pMid=head;
             while(pMid->next!=NULL && pMid->next->val>=x) pMid=pMid->next;
             if(pMid->next==NULL)  return head;
            head=pMid->next;
            pMid->next=head->next;
            head->next=pLarge;
            pSmall=head;
        }
        
         while( true)
        {
            pMid=pSmall;
            
             while(pMid->next!=NULL && pMid->next->val>=x) pMid=pMid->next;
             if(pMid->next==NULL)  break;
            
            pLarge=pMid->next;
            pMid->next=pLarge->next;
            pLarge->next=pSmall->next;
            pSmall->next=pLarge;
            
             while(pSmall->next!=NULL && pSmall->next->val<x) pSmall=pSmall->next;
             if(pSmall->next==NULL)  break;
        }
         return head;
    }
};