[Leetcode] 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.

No words!

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *partition(ListNode *head, int x) {
12         if (head == NULL) return head;
13         ListNode *h = new ListNode(-1);
14         h->next = head;
15         ListNode *pre = h, *cur = h, *tmp;
16         while (cur != NULL && cur->next != NULL) {
17             if (cur->next->val < x && cur != pre) {
18                 tmp = cur->next;
19                 cur->next = tmp->next;
20                 tmp->next = pre->next;
21                 pre->next = tmp;
22                 pre = pre->next;
23             } else if (cur->next->val < x && cur == pre) {
24                 pre = pre->next;
25                 cur = cur->next;
26             } else {
27                 cur = cur->next;
28             }
29         }
30         return h->next;
31     }
32 };