In the Leetcode Remove Duplicates from Sorted List problem solution 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.
Solution in C Programming
struct ListNode* deleteDuplicates(struct ListNode* head){
struct ListNode* strobe;
strobe=head;
if(head==NULL){return NULL;}
while(strobe->next!=NULL){
if(strobe->val==strobe->next->val){
strobe->next=strobe->next->next;
continue;
}
strobe=strobe->next;
}
return head;
}Solution in C++ Programming
/**
* 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* temp=head;
while(temp and temp->next)
{
if(temp->val==temp->next->val)
temp->next=temp->next->next;
else
temp=temp->next;
}
return head;
}
};Solution in Java Programming
class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode dummy = head;
while(dummy != null && dummy.next != null){
if (dummy.val == dummy.next.val)
dummy.next = dummy.next.next;
else
dummy = dummy.next;
}
return head;
}
}Solution in Python Programming
class Solution:
def deleteDuplicates(self, head):
ref = ListNode()
p1=head
ref.next=p1
while p1:
if p1.next:
if p1.next.val == p1.val:
p1.next = p1.next.next
else:
p1=p1.next
else:
p1 = p1.next
return ref.nextSolution in C# Programming
/**
* 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) {
var currentNode = head;
while(currentNode != null && currentNode.next != null)
{
if(currentNode.val == currentNode.next.val)
currentNode.next = currentNode.next.next;
else
currentNode = currentNode.next;
}
return head;
}
}Also read,