# 83. Remove Duplicates from Sorted List

## Description

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.

Example 1:

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


Example 2:

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.

## Solutions

Solution 1: Single Pass

We use a pointer $cur$ to traverse the linked list. If the element corresponding to the current $cur$ is the same as the element corresponding to $cur.next$, we set the $next$ pointer of $cur$ to point to the next node of $cur.next$. Otherwise, it means that the element corresponding to $cur$ in the linked list is not duplicated, so we can move the $cur$ pointer to the next node.

After the traversal ends, return the head node of the linked list.

The time complexity is $O(n)$, where $n$ is the length of the linked list. The space complexity is $O(1)$.

• /**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode() {}
*     ListNode(int val) { this.val = val; }
*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
while (cur != null && cur.next != null) {
if (cur.val == cur.next.val) {
cur.next = cur.next.next;
} else {
cur = cur.next;
}
}
}
}

• /**
* 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:
while (cur != nullptr && cur->next != nullptr) {
if (cur->val == cur->next->val) {
cur->next = cur->next->next;
} else {
cur = cur->next;
}
}
}
};

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
while cur and cur.next:
if cur.val == cur.next.val:
cur.next = cur.next.next
else:
cur = cur.next


• /**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
for cur != nil && cur.Next != nil {
if cur.Val == cur.Next.Val {
cur.Next = cur.Next.Next
} else {
cur = cur.Next
}
}
}

• /**
* function ListNode(val) {
*     this.val = val;
*     this.next = null;
* }
*/
/**
* @return {ListNode}
*/
var deleteDuplicates = function (head) {
while (cur && cur.next) {
if (cur.next.val === cur.val) {
cur.next = cur.next.next;
} else {
cur = cur.next;
}
}
};


• /**
* public class ListNode {
*     public int val;
*     public ListNode next;
*     public ListNode(int val=0, ListNode next=null) {
*         this.val = val;
*         this.next = next;
*     }
* }
*/
public class Solution {
while (cur != null && cur.next != null) {
if (cur.val == cur.next.val) {
cur.next = cur.next.next;
} else {
cur = cur.next;
}
}
}
}

• // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
pub fn delete_duplicates(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
let mut dummy = Some(Box::new(ListNode::new(i32::MAX)));
let mut p = &mut dummy;

while let Some(mut node) = current {
current = node.next.take();
if p.as_mut().unwrap().val != node.val {
p.as_mut().unwrap().next = Some(node);
p = &mut p.as_mut().unwrap().next;
}
}
dummy.unwrap().next
}
}