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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

有點用到了倒置鏈表II的方法，將符合要求的結點放置在指向pre指針的後面。這道題的思路應該是找到第一個大於等於x值的結點，他前一個位置始終定位pre指針，放置比x小的結點。

方法一（C++）

 1 class Solution {
 2 public:
 3     ListNode* partition(ListNode* head, int x) {
 4         ListNode* dummy=new ListNode(-1);
 5         dummy->next=head;
 6         ListNode* pre=dummy,* cur=head;
 
 7         while(pre->next&&pre->next->val<x)
 8             pre=pre->next;
 9         cur=pre;
10         while(cur->next){
11             if(cur->next->val<x){
12                 ListNode* t=cur->next;
13                 cur->next=t->next;
14                 t->next=pre->next;
 
15                 pre->next=t;
16                 pre=t;
17             }
18             else
19                 cur=cur->next;
20         }
21         return dummy->next;
22     }
23 };

LeetCode 86. Partition List 劃分鏈表