Problem

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.

My Solution

 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode h = new ListNode(Integer.MIN_VALUE);
        h.next = head;
        ListNode current = head;
        ListNode t = h;
        boolean flag = false;
        while (current != null && current.next != null) {
            if (current.next.val == current.val) {
                current = current.next;
                t.next = current;
                flag = true;
            } else if (flag == true) {
                current = current.next;
                t.next = current;
                if (current.next != null && current.next.val != current.val) {
                    t = t.next;
                    current = current.next;
                }
                flag = false;
            } else if (flag == false) {
                current = current.next;
                t = t.next;
            }
        }
        if (flag == true) {
            current = current.next;
            t.next = current;
            t = t.next;
        } else if (flag == false && current != null) {
            current = current.next;
            t = t.next;
        }
        return h.next;
    }
}

Great Solution

public ListNode deleteDuplicates(ListNode head) {
        if(head==null) return null;
        ListNode FakeHead=new ListNode(0);
        FakeHead.next=head;
        ListNode pre=FakeHead;
        ListNode cur=head;
        while(cur!=null){
            while(cur.next!=null&&cur.val==cur.next.val){
                cur=cur.next;
            }
            if(pre.next==cur){
                pre=pre.next;
            }
            else{
                pre.next=cur.next;
            }
            cur=cur.next;
        }
        return FakeHead.next;
}