lintcode: 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->null and x = 3, return
1->2->2->4->3->5->null.

这题思路貌似很简单。
1.开个两个空间，分别作为小于x和大于等于x的头结点（不放数据）head1,head2。
2.但是就这么简单的程序我却错了好几次，原因：
（1）p1->next=NULL; 没有对新链的末尾next指针赋为NULL，这样链不知道什么时候截止；
（2）p1->next=NULL; 后发现程序还是出错。发现了原因，原来，修改了p1->next也就相当于修改为了p，p修改了p的next就丢失了，所以事先要记录next。

/** * Definition of ListNode * class ListNode { * public: *     int val; *     ListNode *next; *     ListNode(int val) { *         this->val = val; *         this->next = NULL; *     } * } */class Solution {public:    /**     * @param head: The first node of linked list.     * @param x: an integer     * @return: a ListNode      */    ListNode *partition(ListNode *head, int x) {        // write your code here        if(head==NULL){            return NULL;        }        ListNode *head1=new ListNode(0);        ListNode *p1=head1;        ListNode *head2=new ListNode(0);        ListNode *p2=head2;        ListNode* p=head;        while(p){            ListNode* next=p->next;//!!            if(p->val<x){                p1->next=p;                p1=p1->next;                p1->next=NULL;//!!            }else{                p2->next=p;                p2=p2->next;                p2->next=NULL;//!!            }            p=next;        }        p1->next=head2->next;        return head1->next;    }};