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Leetcode-Partition List

时间:2014-11-16 08:15:41      阅读:222      评论:0      收藏:0      [点我收藏+]

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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.

Analysis:

The first element that is not less than x is the point of partition. We record this node and its predecessor. Once we encount a node that is less than x later, we insert it between the point and its predecessor, then update the predecssor to this node.

Solution:

 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         if (head==null || head.next==null) return head;
15         ListNode preHead = new ListNode(-1);
16         preHead.next = head;
17         ListNode preStart = null,start=null;
18         ListNode cur = head;
19         ListNode pre = preHead;
20 
21         while (cur.val<x && cur.next!=null){
22             pre = cur;
23             cur = cur.next;
24         }
25 
26         if (cur.next==null) return head;
27 
28         start = cur;
29         preStart = pre;
30         pre = cur;
31         cur = cur.next;
32         while (cur!=null){
33             if (cur.val<x){
34                 pre.next = cur.next;
35                 preStart.next = cur;
36                 cur.next = start;
37                 preStart = cur;
38                 cur = pre.next;
39             } else {
40                 pre = cur;
41                 cur = cur.next;
42             }
43         }
44 
45         return preHead.next;            
46     }
47 }

 

Leetcode-Partition List

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原文地址:http://www.cnblogs.com/lishiblog/p/4100959.html

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