# Description

# Given a sorted linked list, delete all duplicates such that each element appear only once.

Linked list length <= 30000<=30000

Example

Example 1:

Input:

linked list = null

Output:

null

Explanation:

Empty linked list returns null

Example 2:

Input:

linked list = 1->1->2->null

Output:

1->2->null

Explanation:
Delete duplicate 1

Example 3:

Input:

linked list = 1->1->2->3->3->null

Output:

1->2->3->null

Explanation:

Delete duplicate 1 and 3

# 解題思路

這題很直觀的就是如果我當前的 node 和下一個 node 一樣的話，那我就指向下下一個（把它跳過的意思）

public ListNode deleteDuplicates(ListNode head) {

	if(head == null) return null;

	ListNode node = head;

	while (head.next != null){

		if(node.val == node.next.val){

			node.next = node.next.next;

		} else {

			node = node.next;

		}

	}

	return head;

}

# 快慢指針

這種有序的鏈表，可以讓 fast 快指針先走前面，慢指針 slow 走後面，如果 fast 走到重復的節點，就讓 slow 跳過，走到不重複時，讓 slow 走一步，跟上面解法差不多。

public ListNode deleteDuplicates(ListNode head) {

	if(head == null) return null;

	ListNode slow = head, fast = head;

	while (fast != null){

		if(fast.val != slow.val){

			// 讓 slow 等於現在的 fast  

			//e.g. nums[slow] == nums[fast]			

			slow.next = fast;  

			slow = slow.next; // 不重復時可以往前走

		} 

		fast = fast.next;

	}

	return head;

}