Leetcode: Remove Duplicates from Sorted List

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.

遇到的问题：input{1,1,1}, output{1,1}, expected{1}, 原因在于若temp.val==temp.next.val, 则需要temp.next=temp.next.next, 这时候就不要temp=temp.next了

注意停止条件不是temp!=null，而是temp.next!=null

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
if(head==null) return null;
ListNode temp=head;
while(temp.next!=null){
if(temp.val==temp.next.val){
temp.next=temp.next.next;
}
else temp=temp.next;
}
return head;
}
}

维护两个指针，一个指向当前不重复的最后一个元素，一个进行依次扫描，遇到不重复的则更新第一个指针，继续扫描，否则就把前面指针指向当前元素的下一个（即把当前元素从链表中删除）。时间上只需要一次扫描，所以是O(n)，空间上两个额外指针，是O(1)。

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
if (head == null || head.next == null) return head;
ListNode walker = head;
ListNode runner = head.next;
while (walker != null && runner != null) {
if (walker.val == runner.val) {
walker.next = runner.next;
runner = runner.next;
}
else {
walker = walker.next;
runner = runner.next;
}
}
return head;
}
}