83

题目

给定一个已排序的链表的头 head ， 删除所有重复的元素，使每个元素只出现一次 。返回 已排序的链表 。

示例 1：

输入：head = [1,1,2]
输出：[1,2]

示例 2：

输入：head = [1,1,2,3,3]
输出：[1,2,3]

题解

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

82

题目

给定一个已排序的链表的头 head ， 删除原始链表中所有重复数字的节点，只留下不同的数字 。返回 已排序的链表 。

示例 1：

输入：head = [1,2,3,3,4,4,5]
输出：[1,2,5]

示例 2：

输入：head = [1,1,1,2,3]
输出：[2,3]

题解

/*** 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) {//可能会把第一个元素删除，新建哨兵节点ListNode dummy = new ListNode(0,head);ListNode cur = dummy;while (cur.next != null && cur.next.next != null) {int val = cur.next.val; //有两个以上重复元素if (val == cur.next.next.val) {//原始的next的val与删除后next的val比较while (cur.next != null && val == cur.next.val) {cur.next = cur.next.next;}} else {cur = cur.next;}}return dummy.next;}
}