【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

 

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode *partition(ListNode *head, int x) {

         ListNode *p=head;
         ListNode *p0=NULL;
         ListNode *head1=NULL;
         ListNode *head2=NULL;
         while(p!=NULL)
         {
             ListNode *tmp;
             if(p->val<x)
             {
                 tmp=new ListNode(p->val);
                 if(p0==NULL) head1=tmp;
                 else  p0->next=tmp;
                 p0=tmp;
             }

             p=p->next;
         }

         ListNode *head1End=p0;
         p=head;
         p0=NULL;
         while(p!=NULL)
         {
             ListNode *tmp;
             if(p->val>=x)
             {
                 tmp=new ListNode(p->val);
                 if(p0==NULL) head2=tmp;
                 else p0->next=tmp;
                 p0=tmp;
             }
             p=p->next;
         }

         if(head1End!=NULL) head1End->next=head2;
         else head1=head2;

         return head1;
     }
 };

第二种思路，与第一种方法类似，只是不新建链表进行存储

 

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode *partition(ListNode *head, int x) {

         if(head==NULL) return NULL;

         ListNode *p=head;

         ListNode *head1=NULL;
         ListNode *head2=NULL;

         ListNode *p0=NULL;
         ListNode *p1=NULL;

         while(p!=NULL)
         {
             if(p->val<x)
             {
                 if(p0==NULL) head1=p;
                 else  p0->next=p;
                 p0=p;
             }
             else
             {
                 if(p1==NULL) head2=p;
                 else p1->next=p;
                 p1=p;
             }

             p=p->next;
         }

         if(p1!=NULL) p1->next=NULL;
         if(p0!=NULL) p0->next=head2;
         else head1=head2;

         return head1;
     }
 };