# 83. Remove Duplicates

[ **83. Remove Duplicates from Sorted List**](https://leetcode.com/problems/remove-duplicates-from-sorted-list/description/)

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`.\\

&#x20;**My Solutions:**

* 新建dummy记录开头
* 如果head 和head.next 相同，跳过head.next

```
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        
        if (head == null || head.next == null) return head;
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        while (head != null && head.next != null) {
            if (head.val == head.next.val) {
                head.next = head.next.next;
            } else {
                head = head.next;
            }
        }
        return dummy.next;
        
    }
}
```

[ **82. Remove Duplicates from Sorted List II**](https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/description/)

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`.

**My Solutions:**

多增加把重复节点跳过的部分

```
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        
        ListNode dummy = new ListNode(0), slow = dummy, fast = head;
        dummy.next = head;
        while(fast != null) {
    	    while (fast.next != null && fast.val == fast.next.val) {
     		    fast = fast.next;    //while loop to find the last node of the dups.
    	    }
    	    if (slow.next != fast) { //duplicates detected.
    		    slow.next = fast.next; //remove the dups.
    	    } else { //no dup, move down both pointer.
    		    slow = slow.next;
    	    }
    	    fast = fast.next; //reposition the fast pointer.
        }
        return dummy.next;
        
    }
}
```

* recursive

```
        //recursive
         if (head == null) return null;
    
        // if (head.next != null && head.val == head.next.val) {
        //     while (head.next != null && head.val == head.next.val) {
        //         head = head.next;
        //     }
        //     return deleteDuplicates(head.next);
        // } else {
        //     head.next = deleteDuplicates(head.next);
        // }
        // return head;
```
