标题： 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 o  f the two partitions. For example, Given 1->4->3->2->5->2 and x = 3, return 1->2->2->4->3->5. 分析：要注意head1==head2这种疑问。 代码如下： ??????? ListNode *partition(ListNode *head, int x) { ??????? if(head==NULL)return head; ??????? ListNode *head1=head; ??????? if(head->val>=x) ??????? { ??????????? while(head1->next!=NULL