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;
}
}