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.

Analysis: Linked List 惯用套路，Runner Technique(Two Pointers), 一些技巧就是：设置head的前置假节点prev，两个pointer：current和runner都指到这个prev，然后进行判断总是判断 current.next 或者 runner.next. 这样做按照我多次做类似题的经验来说，是最方便省事不容易出错的。这道题一次过。思路就是current 和 runner 一直移动直到找到 current.next >= x 为止，这里就是后面小于x的元素将要插入的位置，current便停在这里，指示这个位置，runner继续往后面寻找，把每一个小于x的元素都插入到current.next 的位置。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode partition(ListNode head, int x) {
14         ListNode prev = new ListNode(-1);
15         prev.next = head;
16         ListNode current = prev;
17         ListNode runner = prev;
18         while (current.next != null && current.next.val < x) {
19             current = current.next;
20             runner = runner.next;
21         }
22         while (runner.next != null) {
23             if (runner.next.val < x) {
24                 ListNode temp = runner.next;
25                 runner.next = runner.next.next;
26                 temp.next = current.next;
27                 current.next = temp;
28                 current = current.next;
29             }
30             else runner = runner.next;
31         }
32         return prev.next;
33     }
34 }

 

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode partition(ListNode head, int x) {
14         ListNode dummy = new ListNode(-1);
15         dummy.next = head;
16         ListNode cur = dummy;
17         ListNode runner;
18         while (cur.next != null && cur.next.val < x) {
19             cur = cur.next;
20         }
21         runner = cur;
22         while (runner.next != null) {
23             if (runner.next.val < x) {
24                 ListNode next = runner.next.next;
25                 runner.next.next = cur.next;
26                 cur.next = runner.next;
27                 cur = cur.next;
28                 runner.next = next;
29             }
30             else runner = runner.next;
31         }
32         return dummy.next;
33     }
34 }