力扣 83.删除排序链表中的重复元素

一、题目

二、思路

需要比较当前节点与前一个节点的 val 是否相同，相同则删除当前节点。（如果是比较当前节点和下一节，需要检查 cur.next 是否为空）

三、代码

/**
 * 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) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode pre = head;
        ListNode cur = head.next;
        while (cur != null) {
            if (cur.val == pre.val) {
                pre.next = cur.next;
            } else {
                pre = cur;
            }
            cur = cur.next;
        }
        return head;
    }
}

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) {
            return head;
        }
        ListNode cur = head;
        while (cur.next != null) {
            if (cur.val == cur.next.val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }
        return head;
    }
}