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.

注意： 1。创建DummyNode。2.记得将尾部节点的next置为null。

 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         if (head == null || head.next == null) {
12             return head;
13         }
14         ListNode leftDummy = new ListNode(0);
15         ListNode rightDummy = new ListNode(0);
16         ListNode left = leftDummy;
17         ListNode right = rightDummy;
18         while (head != null) {
19             if (head.val < x) {
20                 left.next = head;
21                 left = left.next;
22             } else {
23                 right.next = head;
24                 right = right.next;
25             }
26             head = head.next;
27         }
28         left.next = rightDummy.next;
29         right.next = null;
30         return leftDummy.next;
31     }
32 }

 

posted @ 2016-03-24 08:59  YuriFLAG  阅读(181)  评论(0)    收藏  举报