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;
    }
}