原题地址：
https://oj.leetcode.com/problems/partition-list/
题目内容：
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.
方法：
链表的基本功。
维护两个带头结点的链表，一个存 < ，一个存 >=，然后一个连接就可以了。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode partition(ListNode head, int x) {
if (head == null)
return null;
ListNode lessHead = new ListNode(0);
ListNode moreHead = new ListNode(0);
ListNode lessTail = lessHead;
ListNode moreTail = moreHead;
ListNode tmp = null;
while (head != null)
{
tmp = head;
head = head.next;
tmp.next = null;
if (tmp.val < x)
{ lessTail.next = tmp;
lessTail = lessTail.next;
}
else
{
moreTail.next = tmp;
moreTail = moreTail.next;
}
}
if (lessHead.next != null)
lessTail.next = moreHead.next;
else
lessHead.next = moreHead.next;
return lessHead.next;
}
}