Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

思路：这个题在刚開始做的时候想的有点，怎么都没办法正确解出。后面把代码所有删除重写。思路是记录当前节点p=head，然后head往下遍历，当head的值不等于head.next的值时，结束。比較p==head，相等说明没有反复，连接上。不相等说明有反复，跳过就可以。

详细代码例如以下：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {

        ListNode first = new ListNode(0);
        ListNode last = first;
        
        ListNode p = head;
        
        while(head != null){
        	while(head.next != null){//p不动，head后移直到head.next与p不相等
        		if(p.val == head.next.val){
        			head = head.next;//相等循环
        		}else{
        			break;//不相等结束
        		}
        	}
        	if(p == head){//仅仅有一个
        		last.next = p;//加入
        		last = last.next;
        	}
        	p = head = head.next;//有多个则不加入
        	last.next = null;//去掉关联
        }
		return first.next;
    }
}