notes/OJ notes/pages/Leetcode Remove-Duplicates-From-Sorted-List.md

# Leetcode Remove-Duplicates-From-Sorted-List

#### 2022-06-16 14:21

> ##### Data structures:
>
> #DS #linked_list
>
> ##### Difficulty:
>
> #coding_problem #difficulty_easy
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> Initial encounter: 2022-06-16
> 1st. revision: 2022-07-02

##### Related topics:

##### Links:

- [Link to problem](https://leetcode.com/problems/remove-duplicates-from-sorted-list/)

---

### Problem

Given the `head` of a sorted linked list, _delete all duplicates such that each element appears only once_. Return _the linked list **sorted** as well_.

#### Examples

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/01/04/list1.jpg)

**Input:** head = [1,1,2]
**Output:** [1,2]

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/01/04/list2.jpg)

**Input:** head = [1,1,2,3,3]
**Output:** [1,2,3]

#### Constraints

- The number of nodes in the list is in the range `[0, 300]`.
- `-100 <= Node.val <= 100`
- The list is guaranteed to be **sorted** in ascending order.

### Thoughts

This can be implemented using recursion and iteration, like [[Leetcode Reverse-Linked-List]].
Recursion is not recommended, since it will take up O(n) space.
In iteration, one pointer or 2 pointers can be used.

To understand the ptr->next->next = ptr->next, see[[Leetcode Reverse-Linked-List#Solution]]

### Solution

One pointer method

```cpp
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode *deleteDuplicates(ListNode *head) {
    ListNode *ptr = head;
    while (ptr != NULL && ptr->next != NULL) {
      if (ptr->val == ptr->next->val) {
        // Duplicate found
        auto tmp = ptr->next;
        ptr->next = ptr->next->next;
        delete tmp;
        // I done set ptr to ptr->next, in case new ptr->next is duplicated
      } else {
        ptr = ptr->next;
      }
    }
    return head;
  }
};
```