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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.

题目要求对链表分段，所有小于X的元素都排在大于等于X的前面。

代码如下：

class Solution {public:    ListNode* partition(ListNode* head, int x) {        if (head == NULL)			return head;		ListNode tmpHead(0);		tmpHead.next = head;		ListNode *p = &tmpHead;		ListNode *q = p->next;		while(q && q->val
   
    next;			q = p->next;		}		while (q != NULL)		{			ListNode * tmpNode = q->next;			if(tmpNode == NULL)			    break;			if (tmpNode->val < x)			{				q->next = tmpNode->next;				tmpNode->next = p->next;				p->next = tmpNode;				p = p->next;			}else{			    q = q->next;			}		}		return tmpHead.next;    }};
   

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