86.Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greate R than or equal to x.

You should preserve the original relative order of the nodes in each of the.

for example,
1->4->3->2->5->2  and return 1->2->2->4->3->5 .

Idea: The task of this problem is to divide the list, smaller than x of the linked list node is now larger than X or equal to the node in front, but also to maintain the original relative position do not change. We create a new two list of L1,L2, traverse the original linked list, the value is less than X nodes are added to L1, the value is greater than or equal to the X nodes are joined L2, and finally the L2 connected to the back of the L1, the resulting list is the result of our request.

  
 

   
  

  
/**
   
  

  
 * 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 *head1,*head2;
   
  

  
 head1=head2=NULL;
   
  

  
 ListNode *cur1,*cur2;
   
  

  
 cur1=cur2=NULL;
   
  

  
 while(head){
   
  

  
 ListNode *cur=head;
   
  

  
 head=head->next;
   
  

  
 cur->next=NULL;
   
  

  
 if(cur->val<x){
   
  

  
 if(!head1){
   
  

  
 head1=cur;
   
  

  
 cur1=head1;
   
  

  
 }
   
  

  
 else{
   
  

  
 cur1->next=cur; 
   
  

  
 cur1=cur1->next;
   
  

  
 }
   
  

  
 
   
  

  
 }else{
   
  

  
 if(!head2){
   
  

  
 head2=cur;
   
  

  
 cur2=head2;
   
  

  
 }else{
   
  

  
 cur2->next=cur;
   
  

  
 cur2=cur2->next;
   
  

  
 }
   
  

  
 }
   
  

  
 }
   
  

  
 if(!cur1)
   
  

  
 return head2;
   
  

  
 cur1->next=head2;
   
  

  
 return head1;
   
  

  
 
   
  

  
 
   
  

  
 
   
  

  
 
   
  

  
 }
   
  

  
};
  
 

 

From for notes (Wiz)

86.Partition List