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
        // 解题思路：
        // 1.如果head == null || head.next == null,则返回head
        // 2.新增一个虚节点root
        // 采用双指针pre = root,cur = head;
        // 3.新建临时变量temp=Integer.MAX_VALUE
        // 4.遍历链表，如果cur.val = temp,则cur = cur.next;
        // 5.如果不相等，则判断cur.val == cur.next.val，相等则 temp = cur.val
        // 6.不等则 pre.next = cur;pre = cur;temp=cur.val;cur = cur.next
        // 7.最后记得把pre.next 指向null

        if (head == null || head.next == null) {
            return head;
        }

        ListNode root = new ListNode(-1);
        ListNode pre = root;
        ListNode cur = head;
        int temp = Integer.MAX_VALUE;

        while (cur != null) {
            if (temp == cur.val) {
                cur = cur.next;
            } else {
                if (cur.next != null && cur.val == cur.next.val) {
                    temp = cur.val;
                } else {
                    pre.next = cur;
                    pre = cur;
                    temp = cur.val;
                    cur = cur.next;
                }
            }
        }

        pre.next = null;

        return root.next;

    }
}