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 = nextclassSolution:defdeleteDuplicates(self,head:Optional[ListNode])->Optional[ListNode]:dummy=pre=ListNode(next=head)cur=headwhilecur:whilecur.nextandcur.next.val==cur.val:cur=cur.nextifpre.next==cur:pre=curelse:pre.next=cur.nextcur=cur.nextreturndummy.next
/** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */funcdeleteDuplicates(head*ListNode)*ListNode{dummy:=&ListNode{Next:head}pre,cur:=dummy,headforcur!=nil{forcur.Next!=nil&&cur.Next.Val==cur.Val{cur=cur.Next}ifpre.Next==cur{pre=cur}else{pre.Next=cur.Next}cur=cur.Next}returndummy.Next}