82. Remove Duplicates from Sorted List II

Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

使用队列暂存遍历的节点。
初始化prev为一个dummy节点。
如果当前节点不等于队列里节点的值，则倾倒出队列里的值。如果队列此时只有一个值，则将其添加到prev.next。
遍历完毕后如果队列内只有一个值则将其设置到prev.next。
最后返回dummy.next。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        Queue<ListNode> q = new LinkedList();
        ListNode dummy = new ListNode(0);
        ListNode prev = dummy;
        ListNode curr = head; 
        while(curr != null){
            if(q.isEmpty() || curr.val == q.peek().val){
                q.offer(curr);
                curr = curr.next;
            }
            else{
                if(q.size()==1){
                    prev.next = q.poll();
                    prev = prev.next;
                    prev.next = null;
                }
                else{
                    q.clear();   
                }
            }
        }
        
        if(q.size()==1){
            prev.next = q.poll();
        }
        
        return dummy.next;
    }

}