题目链接：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 and x = 3,

return 1->2->2->4->3->5.

这道题的要求是将链表中小于x的节点放到左侧，大于等于x的节点放到右侧，要保持原有的相对顺序。

思路比较简单，用l标识左侧小于x的节点的最后位置，则每次右移r，当r指向的节点小于x时，把其放到l后面即可。

时间复杂度：O(n)

空间复杂度：O(1)

 1 class Solution
 2 {
 3 public:
 4     ListNode *partition(ListNode *head, int x)
 5     {
 6         ListNode *h = new ListNode(0);
 7         h -> next = head;
 8         
 9         ListNode *l = h, *r = h;
10         while(r -> next != NULL)
11         {
12             if(r -> next -> val < x)
13             {
14                 if(l != r)
15                 {
16                     ListNode *tempr = r -> next -> next;
17                     r -> next -> next = l -> next;
18                     l -> next = r -> next;
19                     l = r -> next;
20                     r -> next = tempr;
21                 }
22                 else
23                 {
24                     l = l -> next;
25                     r = r -> next;
26                 }
27             }
28             else
29                 r = r -> next;
30         }
31         
32         return h -> next;
33     }
34 };