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.

分析：可以将小于x的node放在一个链表中，大于等于x的node放在另一个链表中，然后将两个链表首尾相连。

class Solution {public:    ListNode *partition(ListNode *head, int x) {        if(head == NULL) return head;                ListNode *shead = new ListNode(-1);        ListNode *bhead = new ListNode(-1);                ListNode *ps = shead, *pb = bhead;        while(head){            ListNode *tmp = head;            head = head->next;            tmp->next = NULL;            if(tmp->val < x){                ps->next = tmp;                ps = ps->next;            }else{                pb->next = tmp;                pb = pb->next;            }        }        ps->next = bhead->next;                return shead->next;    }};