[LeetCode] 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)
             return head;
         ListNode *p = head;
         ListNode *lessHead = NULL,*greaterHead =NULL,*pLess=NULL,*pGreater=NULL;
         while(p){
             if(p->val < x ){
                 if(lessHead == NULL){
                   lessHead = p;
                   pLess = p;
                 }else{
                    pLess->next = p;
                    pLess = pLess->next;
                 }
             }else{
                 if(greaterHead == NULL){
                    greaterHead = p ;
                    pGreater  = p;
                 }else{
                    pGreater->next = p;
                    pGreater = pGreater->next;
                 }
             }
             p = p->next;
         }//end while
         if(pGreater!=NULL)
             pGreater->next = NULL;
         if(pLess !=NULL){
              pLess->next = greaterHead;
              return lessHead;
         }else{
              return greaterHead;
         }
           
    }//end func
    
};