方法一：不使用额外的存储空间，直接在原始链表上进行操作。首先用一个指针指向链表头节点开始，然后遍历其后面的节点，将与该指针所指节点数据相同的节点删除。然后将该指针后移一位，继续上述操作。直到该指针移到链表。 void delete_duplicate1(node* head){     node*pPos=head->next;     node*p,*q;     while(pPos!=NULL){//用pPos指针来指示当前移动到什么位置了         p=pPos;        q=pPos->next;        while(q!=NULL){//遍历pPos后面的所有节点，找出节点值与pPos所指节点相同的节点，将其删除             if(pPos->data==q->data){                 node*pDel=q;                 p->next=q->next;                 q=p->next;                 free(pDel);                 }             else{                 p=q;                 q=q->next;                 }             }         pPos=pPos->next;         } }

void delete_duplicate2(node* head) {     node*p=head->next;     node*q=p->next;     memset(hash,0,sizeof(hash));     hash[p->data]=1;//置为1，表示该数已经出现过     while(q!=NULL){         if(hash[q->data]!=0){             node*pDel=q;             p->next=q->next;             q=p->next;             free(pDel);             }         else{             hash[q->data]=1;//置为1，表示该数已经出现过             p=q;             q=q->next;             }         } }

public static void deleteDups(LinkedListNode n) {   Hashtable table = new Hashtable();   LinkedListNode previous = null;   while (n != null) {     if (table.containsKey(n.data)) previous.next = n.next;     else {       table.put(n.data, true);       previous = n;     }     n = n.next;   } } public static void deleteDups2(LinkedListNode head) {     if (head == null) return;     LinkedListNode previous = head;     LinkedListNode current = previous.next;     while (current != null) {       LinkedListNode runner = head;       while (runner != current) { // Check for earlier dups         if (runner.data == current.data) {           LinkedListNode tmp = current.next; // remove current           previous.next = tmp;           current = tmp; // update current to next node           break; // all other dups have already been removed         }         runner = runner.next;       }       if (runner == current) { // current not updated - update now         previous = current;         current = current.next;       }     }  }