Partition List

题目描述：

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

.
解题思路：
模仿快排的一次排序过程即可。

注：在链表的开头加一个虚拟的节点（dummyHead）可以简化处理过程。

AC代码如下：

class Solution {
public:
ListNode* partition(ListNode* head, int x) {
if (head == NULL) return NULL;
ListNode* dummyHead1 = new ListNode(-1);
ListNode* p = dummyHead1;
ListNode* dummyHead2 = new ListNode(-1);
dummyHead2->next = head;
ListNode* q = dummyHead2;
while (q->next != NULL){
if (q->next->val < x){
ListNode* tmp = q->next;
q->next = tmp->next;
p->next = tmp;
p = p->next;
}
else{
q = q->next;
}
}
p->next = dummyHead2->next;
return dummyHead1->next;
}
};