题解 | 删除有序链表中重复的元素-II

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    public ListNode deleteDuplicates (ListNode head) {
        // write code here
        if(head == null) return head;
        Map<Integer,Integer> mp = new HashMap<>();
        ListNode cur = head;
        while(cur != null){
            if(mp.containsKey(cur.val)){
                mp.put(cur.val,(int)mp.get(cur.val)+1);
            }
            else mp.put(cur.val,1);
            cur = cur.next;
        }
        ListNode res = new ListNode(0);
        res.next = head;
        cur = res;
        while(cur.next != null){
            if(mp.get(cur.next.val)!=1) cur.next = cur.next.next;
            else cur = cur.next;
        }
        return res.next;
    }
}

这个是用map来统计出现次数之后跳过节点 更容易理解 但是比之前那个解法稍微复杂了一点

用数组也可以