Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given 1->1->2
, return 1->2
.
Given 1->1->2->3->3
, return 1->2->3
.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */
//思想同昨天的从有序数组去重复元素是一样的 public class Solution { public ListNode deleteDuplicates(ListNode head) { if (head==null || head.next==null) return head; ListNode pre = head; ListNode cur = head.next; //cur与pre用来在list中找重复 ListNode non_dup_end = head; //用来标记无重复list的最后一个节点 while(cur != null){ if (pre.val != cur.val){ //如果不是重复的话,属正常态,就把non_dup_end向后移一位,然后把cur.val赋过去 non_dup_end = non_dup_end.next; non_dup_end.val = cur.val; } cur = cur.next; //如果是重复的话,本次循环什么都不做,直到找到后边第一个不再重复的元素值赋给non_dup_end pre = pre.next; } non_dup_end.next = null; //不同于数组,循环之后要记得把无重复list后边多余的节点删掉 return head; } }