86.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.

思路：此题的任务是划分链表，比x小的链表节点都出现在比x大或者等于的节点前面，而且还要保持原来的相对位置不要发生变化。我们新建两个链表了l1,l2，遍历原来的链表 ，把值小于x的节点都加入l1,把值大于或者等于x的节点都加入l2，最后将l2连接至l1的后面，所得的链表就是我们所求的结果了。

1. /**
    * 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;
           
           
           
           
       }
   };