leetcode86 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.

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* partition(ListNode* head, int x) {
12         ListNode *ans=new ListNode(0);
13         ListNode *ansp=ans;
14         
15         ListNode *p=head;
16         while(p)
17         {
18             if(p->val<x)
19             {
20                 ListNode *temp=new ListNode(p->val);
21                 ansp->next=temp;
22                 ansp=ansp->next;
23             }
24             p=p->next;
25         }
26         p=head;
27         while(p)
28         {
29             if(p->val>=x)
30             {
31                 ListNode *temp=new ListNode(p->val);
32                 ansp->next=temp;
33                 ansp=ansp->next;
34             }
35             p=p->next;
36         }
37         return ans->next;
38     }
39 };
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posted @ 2016-01-06 16:05  西小贝  阅读(120)  评论(0)  编辑  收藏  举报