[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.

思路：新建两个头结点，分别把小于x 的节点链接到第一个头结点后，把大于或等于x 的节点链接到第二个头节点后。

　　　　时间复杂度O(n),空间复杂度O(1)

 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         
14         ListNode *plow = new ListNode(-1);
15         ListNode *plhead = plow;
16         ListNode *phigh = new ListNode(-1);
17         ListNode *phhead = phigh;
18         ListNode *pt = head;
19         
20         while (pt != NULL) {
21             if (pt->val < x) {
22                 plow->next = pt;
23                 plow = pt;
24             } else {
25                 phigh->next = pt;
26                 phigh = pt;
27             }
28             pt = pt->next;
29         }
30         
31         plow->next = phhead->next;
32         phigh->next = NULL;
33         
34         return plhead->next;
35     }
36 };