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题目描述：

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：

遍历一次链表即可，每遍历一个元素比较其是否与它的下一个元素相等。若是，指针指向下下个元素；否则继续遍历下一个元素。

代码实现：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null){
            return null;
        }
        ListNode p = head;
        while(p.next!=null){
            if(p.val == p.next.val){
                p.next = p.next.next;
            }else{
                p = p.next;
            }
        }
        return head;
    }
}

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