算法思想：

两个指针，s将小于x的串起来，e将大于等于x的串起来，最后将s的next指向e，注意处理head指针就好了

/**
 * 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)return NULL;
        ListNode *p,*q,*e,*s,*start,*end;
        start=s=new ListNode(0);
        end=e=new ListNode(0);
        p=head;
        while(p){
            if(p->val<x){
                s->next=p;
                s=p;
            }
            else{
                e->next=p;
                e=p;
            }
            p=p->next;
        }
        e->next=NULL;
        s->next=end->next;
        return start->next;
    }
};

LeetCode OJ:Partition List