82. Remove Duplicates from Sorted List II

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.

/**
 * 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 || head.next == null) 
            return head;  
        ListNode newHead = new ListNode(-1);  
        newHead.next = head;  
        ListNode p1 = newHead;  
        ListNode p2 = head;  
        while(p2 != null){  
            boolean dup = false;  
            while(p2.next != null && p2.val == p2.next.val){  
                dup = true;  
                p2 = p2.next;  
            }  
            if(dup){  
                p2 = p2.next;  
                continue;  
            }  
            p1.next = p2;  
            p1 = p1.next;  
            p2 = p2.next;  
        }  
        p1.next = p2;  
        return newHead.next;  
    }
}

/**
 * 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 l1 = new ListNode(0);
        ListNode l2, left, right;
        l1.next = l2 = head;
        left = l1;

        int value = 1;
        while(l2.next != null) {
            right = l2;
            l2 = l2.next;
            if(l2.val == right.val) value++;
            else {
                if(value < 2) {
                    left.next = right;
                    left = right;
                }
                value = 1;
            }
        }
        if(value < 2) left.next = l2;
        else left.next = null;
        return l1.next;
    }
}