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; } }
【原创】leetCodeOj ---Partition List 解题报告
原文：http://www.cnblogs.com/shadowmydx/p/4123814.html