Problem: 83. 删除排序链表中的重复元素

题目描述

思路

1.定义快慢指针fast、slow均指向head；
2.每次fast后移一位，当fast和slow指向的节点值不一样时，将slow.next指向fast同时使slow指向fast；
3.去除链表尾部重复元素：最后判断slow是否指向fast，若没有则slow及其后面的元素均是重复的，直接使得slow.next为空即可

复杂度

时间复杂度:

$O(n)$;其中$n$是链表的原始长度

空间复杂度:

$O(1)$;

Code

/*** 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 {/*** Remove Duplicates from Sorted List** @param head The head node of sorted linked list* @return ListNode*/public ListNode deleteDuplicates(ListNode head) {ListNode slow = head;ListNode fast = head;while (fast != null) {if (slow.val != fast.val) {slow.next = fast;slow = fast;}fast = fast.next;}//Determine if there are any duplicate elements at the endif (slow != fast) {slow.next = null;}return head;}
}