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.

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        if(head == NULL ||  head->next == NULL) return head;
        ListNode* p =  new ListNode(), *pre_p = p;
        ListNode *q =  new ListNode(), *pre_q = q;
        while(head){
            if(head->val < x){
                ListNode *tmp = head->next;
                head->next = pre_p->next;
                pre_p->next = head;
                pre_p = pre_p->next;
                head=tmp;
            }else{
                ListNode *tmp = head->next;
                head->next = pre_q->next;
                pre_q->next = head;
                pre_q = pre_q->next;
                head=tmp;
            }
        }
        pre_p->next = q->next;
        return p->next;
    }
};