Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given 1->2->3->3->4->4->5
, return 1->2->5
.
Given 1->1->1->2->3
, return 2->3
.
具体代码如下:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode deleteDuplicates(ListNode head) { ListNode first = new ListNode(0); ListNode last = first; ListNode p = head; while(head != null){ while(head.next != null){//p不动,head后移直到head.next与p不相等 if(p.val == head.next.val){ head = head.next;//相等循环 }else{ break;//不相等结束 } } if(p == head){//只有一个 last.next = p;//添加 last = last.next; } p = head = head.next;//有多个则不添加 last.next = null;//去掉关联 } return first.next; } }
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leetCode 82.Remove Duplicates from Sorted List II (删除排序链表的重复II) 解题思路和方法
原文地址:http://blog.csdn.net/xygy8860/article/details/47002063