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Partition List

时间:2016-07-03 08:10:42      阅读:232      评论:0      收藏:0      [点我收藏+]

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.

Example

Given 1->4->3->2->5->2->null and x = 3,
return 1->2->2->4->3->5->null.

分析:
把nodes分别放在两个不同的list里面,然后合并。
 1 /**
 2  * Definition for ListNode.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int val) {
 7  *         this.val = val;
 8  *         this.next = null;
 9  *     }
10  * }
11  */ 
12 public class Solution {
13     /**
14      * @param head: The first node of linked list.
15      * @param x: an integer
16      * @return: a ListNode 
17      */
18     public ListNode partition(ListNode head, int x) {
19         ListNode left = new ListNode(0);
20         ListNode right = new ListNode(0);
21         
22         ListNode leftHead = left;
23         ListNode rightHead = right;
24         
25         while (head != null) {
26             if (head.val < x) {
27                 left.next = head;
28                 left = head;
29             } else {
30                 right.next = head;
31                 right = head;
32             }
33             head = head.next;
34         }
35         left.next = rightHead.next;
36         right.next = null;
37         return leftHead.next;
38     }
39 }

转载请注明出处:cnblogs.com/beiyeqingteng/

Partition List

原文:http://www.cnblogs.com/beiyeqingteng/p/5636536.html

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