leetcode--Partition List

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.

[java]  view plain  copy

1. /** 
2.  * Definition for singly-linked list. 
3.  * public class ListNode { 
4.  *     int val; 
5.  *     ListNode next; 
6.  *     ListNode(int x) { val = x; } 
7.  * } 
8.  */  
9. public class Solution {  
10.     public ListNode partition(ListNode head, int x) {  
11.         ListNode first = new ListNode(0);  
12.         ListNode second = new ListNode(0);  
13.         ListNode res = first;  
14.         ListNode res2 = second;  
15.         ListNode cur = head;  
16.         while(cur!=null){  
17.             if(cur.val<x){  
18.                 first.next = cur;  
19.                 cur = cur.next;  
20.                 first.next.next = null;  
21.                 first = first.next;  
22.             }else{  
23.                 second.next = cur;  
24.                 cur = cur.next;  
25.                 second.next.next = null;  
26.                 second = second.next;  
27.             }  
28.         }  
29.         first.next = res2.next;  
30.         return res.next;  
31.     }  
32. }  

原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/46429569