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.

Solution:

首先用hashmap计算每个元素出现的次数，然后给list加个头指针，然后遍历链表，将次数多于1的都删掉。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        // Start typing your Java solution below
        // DO NOT write main() function
        HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
        ListNode cur = head;
        while(cur != null){
            if(map.containsKey(cur.val)){
                map.put(cur.val,map.get(cur.val) + 1);
            }else{
                map.put(cur.val,1);
            }
            cur = cur.next;
        }
        ListNode header = new ListNode(-1);
        header.next = head;
        cur = header;
        while(cur.next != null){
            if(map.get(cur.next.val) > 1){
                cur.next = cur.next.next;
            }
            else{
                cur = cur.next;
            }
        }
        return header.next;
    }
}