题解 | #删除有序链表中重复的元素-I#

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.判断head == null || head.next == null,则返回head,因为最多一个节点不可能重复,
        // 增加一个虚节点root
        // 2.采用双指针，cur = head 和 next = head.next,val = cur.val
        // 3.遍历链表，如果val == next.val,则说明值相同，则将next往后面移动，next = next.next
        // 4.如果不相等,则cur.next = next;cur = next;val = next.val;next = next.next
        // 5.遍历完链表，记得将cur.next 指向null,因为很可能后面的都相同
        if (head == null || head.next == null) {
            return head;
        }
        ListNode root = new ListNode(-1);
        root.next = head;

        ListNode cur = head;
        ListNode next = cur.next;
        int val = cur.val;

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

        cur.next = next;


        return root.next;
    }
}