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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.
思路：找出以x为划分的两段链表，并连接在一起，注意维护链表头。
代码：
public class Solution { public ListNode partition(ListNode head, int x) { ListNode small = new ListNode(0); //两个新链表 ListNode big = new ListNode(0); ListNode smallTemp = small; //保存链表头 ListNode bigTemp = big; while(head!=null){ if(head.val < x){ big.next = head; big = big.next; }else{ small.next = head; small = small.next; } head = head.next; } big.next = smallTemp.next; //连接 small.next = null; return bigTemp.next; } }
[LeetCode-JAVA] Partition List
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原文地址：http://www.cnblogs.com/TinyBobo/p/4512567.html
