移除链表中重复元素

【链表】

Q:Write code to remove duplicates from an unsorted linked list 
     FOLLOW UP 
     How would you solve this problem if a temporary buffer is not allowed?

题目：编码实现从无序链表中移除重复项。

         如果不能使用临时缓存，你怎么编码实现？

解答：

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;
        }
}

方法二：如果允许使用额外的空间，则能通过空间换时间，来降低算法的复制度。可以使用hash表来完成，既然是面试题，我们这里可以暂时先不考虑使用hash可能带来的一些问题，先假设它是完美的。即假设它能将任意整数hash到一定范围，不会出现负数下标，不会出现hash冲突等。

JAVA参考代码：

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;
      }
    }
 }

作者：Viidiot   微信公众号：linux-code

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;
      }
    }
 }