【Leetcode】Remove Duplicates from Sorted List in JAVA

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,

Given

1->1->2

, return

1->2

.

Given

1->1->2->3->3

, return

1->2->3

.

思路非常easy。因为乖乖的sort好了，就是推断下一个是不是比它大就好了，假设大，那么跳过下一个直接link到下一个的下一个。可是此时注意。考虑假设是1->1->1这样的情况。当你把第二个1删掉之后。指针一定要保留在第一个的位置，这样才干够接着推断这个1与再下一个1是不是相等（即第一个1和第3个1）。

唯一须要格外注意的情况就是最后两个，因为你要p.next=p.next.next来删除，所以对于最后两个不存在next的next。所以直接等于null就好啦

package testAndfun;

public class deleteDuplicates {
public static void main(String args[]){
deleteDuplicates dp = new deleteDuplicates();
ListNode head  = new ListNode(3);
ListNode p1 = new ListNode(3);
head.next=p1;
ListNode p2  = new ListNode(3);
p1.next =  p2;
ListNode p3 = new ListNode(3);
p2.next = p3;
ListNode p4 = new ListNode(13);
p3.next = p4;
prinf(head);
prinf(dp.deleteDup(head));
}

private static void prinf(ListNode input){
while(input!=null)	{
System.out.print(input.val+"->");
input = input.next;
}
System.out.println();
}

public ListNode deleteDup(ListNode head){
if(head==null||head.next==null)	return head;//if no head, what should I do?
ListNode p=head;
int i=0;
//System.out.println(p.val+" and "+p.next.val);
while(p.next != null){
if(p.val==p.next.val&&p.next.next!=null)	{
//System.out.println("go first"+p.val);//"^^"+p.next.val+"%%"+p.next.next.val);
p.next=p.next.next;
continue;//if this and next equal, we should stay in this in case next.next is equal this
}
else if(p.val==p.next.val&&p.next.next==null)	{
//System.out.println("go second"+p.val);
p.next=null;
continue;
}
//System.out.println(p.val+" round "+i++);
p=p.next;
if(p==null)	break;
}
//System.out.print(head.val);
return head;
}

}