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.

Solution:

稍微修改上一个代码即可，遇到次数多于1的，先输出一个，然后将hashmap中标记一下（这里我就是将其value改为0），后面就不输出它即可。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode deleteDuplicates(ListNode head) {
14         // Start typing your Java solution below
15         // DO NOT write main() function
16           HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
17         ListNode cur = head;
18         while(cur != null){
19             if(map.containsKey(cur.val)){
20                 map.put(cur.val,map.get(cur.val) + 1);
21             }else{
22                 map.put(cur.val,1);
23             }
24             cur = cur.next;
25         }
26         ListNode header = new ListNode(-1);
27         header.next = head;
28         cur = header;
29         while(cur.next != null){
30             if(map.get(cur.next.val) > 1){
31                 map.put(cur.next.val,0);
32                 cur = cur.next;
33             }
34             else if(map.get(cur.next.val) == 1){
35                 cur = cur.next;
36             }else{
37                 cur.next = cur.next.next;
38             }
39         }
40         return header.next;
41     }
42 }

我发现可以简化。。。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode deleteDuplicates(ListNode head) {
14         // Start typing your Java solution below
15         // DO NOT write main() function
16         HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
17         ListNode header = new ListNode(-1);
18         header.next = head;
19         ListNode cur = header;
20         while(cur.next != null){
21             if(map.containsKey(cur.next.val)){
22                 cur.next = cur.next.next;
23             }else{
24                 map.put(cur.next.val,1);
25                 cur = cur.next;
26             }
27         }
28         return header.next;
29     }
30 }

第二遍：由于linkedlist 已经被sorted，所以可以直接用双指针来完成。使用constant extra space。

 1 public class Solution {
 2     public ListNode deleteDuplicates(ListNode head) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int value = Integer.MAX_VALUE;
 6         ListNode cur = head, per = head;
 7         while(per != null){
 8             if(per.val != value){
 9                 value = per.val;
10                 cur = per;
11                 per = per.next;
12             }else{
13                 per = per.next;
14                 cur.next = per;
15             }
16         }
17         return head;
18     }
19 }