【leetcode】Partition List

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.

 
最简答的，扫描两次链表
第一次找到所有比x小的节点，并保存下来
第二次找到所有比x大的节点，并保存下来
然后把他们两连起来
 
  
 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        
13         ListNode *p=head;
14         ListNode *p0=NULL;
15         ListNode *head1=NULL;
16         ListNode *head2=NULL;
17         while(p!=NULL)
18         {
19             ListNode *tmp;
20             if(p->val<x)
21             {
22                 tmp=new ListNode(p->val);
23                 if(p0==NULL) head1=tmp;
24                 else  p0->next=tmp;
25                 p0=tmp;
26             }
27            
28             p=p->next;
29         }
30        
31         ListNode *head1End=p0;
32         p=head;
33         p0=NULL;
34         while(p!=NULL)
35         {
36             ListNode *tmp;
37             if(p->val>=x)
38             {
39                 tmp=new ListNode(p->val);
40                 if(p0==NULL) head2=tmp;
41                 else p0->next=tmp;
42                 p0=tmp;
43             }
44             p=p->next;
45         }
46        
47        
48         if(head1End!=NULL) head1End->next=head2;
49         else head1=head2;
50        
51         return head1;
52     }
53 };

 

第二种思路，与第一种方法类似，只是不新建链表进行存储
 
 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        
13         if(head==NULL) return NULL;
14        
15         ListNode *p=head;
16        
17         ListNode *head1=NULL;
18         ListNode *head2=NULL;
19        
20         ListNode *p0=NULL;
21         ListNode *p1=NULL;
22        
23        
24        
25         while(p!=NULL)
26         {
27             if(p->val<x)
28             {
29                 if(p0==NULL) head1=p;
30                 else  p0->next=p;
31                 p0=p;
32             }
33             else
34             {
35                 if(p1==NULL) head2=p;
36                 else p1->next=p;
37                 p1=p;
38             }
39            
40             p=p->next;
41         }
42        
43         if(p1!=NULL) p1->next=NULL;
44         if(p0!=NULL) p0->next=head2;
45         else head1=head2;
46        
47         return head1;
48     }
49 };

 

posted @ 2015-01-18 18:34  H5开发技术  阅读(224)  评论(0)  编辑  收藏  举报