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;}