Partition List

题目：

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

.
分析：
意思是比x小的元素按顺序放到x前面，按元素原来的顺序，而不是从小到大排序；



参考代码如下：
http://blog.csdn.net/linhuanmars/article/details/24446871

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode partition(ListNode head, int x) {
if(head == null)
return null;
ListNode helper = new ListNode(0);
helper.next = head;
ListNode walker = helper;
ListNode runner = helper;
while(runner.next!=null)
{
if(runner.next.val<x)
{
if(walker!=runner)
{
ListNode next = runner.next.next;
runner.next.next = walker.next;
walker.next = runner.next;
runner.next = next;
}
else
runner = runner.next;
walker = walker.next;
}
else
{
runner = runner.next;
}
}
return helper.next;
}
}