82. Remove Duplicates from Sorted List II

Description

Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

Example 1:

Input: head = [1,2,3,3,4,4,5]
Output: [1,2,5]


Example 2:

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

First, we create a dummy head node $dummy$, and set $dummy.next = head$. Then we create a pointer $pre$ pointing to $dummy$, and a pointer $cur$ pointing to $head$, and start traversing the linked list.

When the node value pointed by $cur$ is the same as the node value pointed by $cur.next$, we let $cur$ keep moving forward until the node value pointed by $cur$ is different from the node value pointed by $cur.next$. At this point, we check whether $pre.next$ is equal to $cur$. If they are equal, it means there are no duplicate nodes between $pre$ and $cur$, so we move $pre$ to the position of $cur$; otherwise, it means there are duplicate nodes between $pre$ and $cur$, so we set $pre.next$ to $cur.next$. Then we continue to move $cur$ forward. Continue the above operation until $cur$ is null, and the traversal ends.

Finally, return $dummy.next$.

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

• /**
* 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 {
ListNode dummy = new ListNode(0, head);
ListNode pre = dummy;
while (cur != null) {
while (cur.next != null && cur.next.val == cur.val) {
cur = cur.next;
}
if (pre.next == cur) {
pre = cur;
} else {
pre.next = cur.next;
}
cur = cur.next;
}
return dummy.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:
ListNode* dummy = new ListNode(0, head);
ListNode* pre = dummy;
while (cur) {
while (cur->next && cur->next->val == cur->val) {
cur = cur->next;
}
if (pre->next == cur) {
pre = cur;
} else {
pre->next = cur->next;
}
cur = cur->next;
}
return dummy->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:
while cur.next and cur.next.val == cur.val:
cur = cur.next
if pre.next == cur:
pre = cur
else:
pre.next = cur.next
cur = cur.next
return dummy.next


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

• /**
* class ListNode {
*     val: number
*     next: ListNode | null
*     constructor(val?: number, next?: ListNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.next = (next===undefined ? null : next)
*     }
* }
*/

function deleteDuplicates(head: ListNode | null): ListNode | null {
const dummy = new ListNode(0, head);
let pre = dummy;
while (cur) {
while (cur.next && cur.val === cur.next.val) {
cur = cur.next;
}
if (pre.next === cur) {
pre = cur;
} else {
pre.next = cur.next;
}
cur = cur.next;
}
return dummy.next;
}


• /**
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @return {ListNode}
*/
var deleteDuplicates = function (head) {
const dummy = new ListNode(0, head);
let pre = dummy;
while (cur) {
while (cur.next && cur.val === cur.next.val) {
cur = cur.next;
}
if (pre.next === cur) {
pre = cur;
} else {
pre.next = cur.next;
}
cur = cur.next;
}
return dummy.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 {
ListNode dummy = new ListNode(0, head);
ListNode pre = dummy;
while (cur != null) {
while (cur.next != null && cur.next.val == cur.val) {
cur = cur.next;
}
if (pre.next == cur) {
pre = cur;
} else {
pre.next = cur.next;
}
cur = cur.next;
}
return dummy.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(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
let mut dummy = Some(Box::new(ListNode::new(101)));
let mut pev = dummy.as_mut().unwrap();
let mut pre = 101;
while let Some(mut node) = cur {
cur = node.next.take();
if node.val == pre || (cur.is_some() && cur.as_ref().unwrap().val == node.val) {
pre = node.val;
} else {
pre = node.val;
pev.next = Some(node);
pev = pev.next.as_mut().unwrap();
}
}
dummy.unwrap().next
}
}