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.
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| # 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]:
dummy = pre = ListNode(next=head)
cur = head
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
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| /**
* Definition for singly-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 {
public ListNode deleteDuplicates(ListNode head) {
ListNode dummy = new ListNode(0, head);
ListNode pre = dummy;
ListNode cur = head;
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;
}
}
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| /**
* 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* dummy = new ListNode(0, head);
ListNode* pre = dummy;
ListNode* cur = head;
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;
}
};
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| /**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func deleteDuplicates(head *ListNode) *ListNode {
dummy := &ListNode{Next: head}
pre, cur := dummy, head
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
}
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| /**
* Definition for singly-linked list.
* 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;
let cur = head;
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;
}
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| // 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 cur = head;
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
}
}
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| /**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var deleteDuplicates = function (head) {
const dummy = new ListNode(0, head);
let pre = dummy;
let cur = head;
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;
};
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| /**
* Definition for singly-linked list.
* 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 {
public ListNode DeleteDuplicates(ListNode head) {
ListNode dummy = new ListNode(0, head);
ListNode pre = dummy;
ListNode cur = head;
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;
}
}
|
