leetCode 82.Remove Duplicates from Sorted List II （删除排序链表的重复II） 解题思路和方法

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：这个题在刚开始做的时候想的有点，怎么都没办法正确解出。后面把代码全部删除重写，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。比较p==head，相等说明没有重复，连接上；不相等说明有重复，跳过即可。
具体代码如下：

/**
* 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){//只有一个
last.next = p;//添加
last = last.next;
}
p = head = head.next;//有多个则不添加
last.next = null;//去掉关联
}
return first.next;
}
}