leetcode------Partition List

标题：	Partition List
通过率：	27.5%
难度：	中等

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.

本题就是利用一个target将链表分成两部分，小于target和大于target的

用两个链表进行记录，small和big进行连接时候一定要将.next置null，否则内存会一直叠加。连接的过程就是指针的重定向过程，所以一定要置null，要不然会讲整个链表连接过来，代码如下：

 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 small=new ListNode(1);

15         ListNode big=new ListNode(1);

16         ListNode s=small,b=big;

17         if(head==null)return head;

18         while(head!=null){

19             if(head.val<x){

20                 s.next=head;

21                 head=head.next;

22                 s=s.next;

23                 s.next=null;

24             }

25             else{

26                 b.next=head;

27                 head=head.next;

28                 b=b.next;

29                 b.next=null;

30             }

31         }

32         s.next=big.next;

33         return small.next;

34     }

35 }