LeetCode_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.

代码：

struct ListNode {
      int val;
      ListNode *next;
      ListNode(int x) : val(x), next(NULL) {}
  };
ListNode* partition(ListNode* head, int x)
{
      if(head==NULL) return NULL;
      ListNode* begin = head;
      ListNode* second = head->next;
      ListNode* second_pre = head;
      while(second!=NULL)
      {
         if(second->val < x)
         {
            if(begin==head&&begin->val >=x)//第一个节点就是比较大的节点 
            {
                    ListNode* head1 = new ListNode(second->val);
                    head1->next = head;
                    head = head1;
                    //删除second指向的节点 
                    second_pre -> next = second -> next;
                    second = second_pre->next;
                    begin = head;
                    
            }
            else if(second_pre->val < x)//连续两个都比较小 
            {
                  second_pre = second;
                  begin = second_pre;
                  second = second->next;
            }
            else
            {
                   
                   second_pre->next = second->next;
                   second->next = begin->next;
                   begin->next = second;
                   second = second_pre->next;
                   begin = begin ->next;
            }
         }
         else
         {
                  second_pre = second;
                  second = second->next;
         }
      } 
      return head;
}