Problem Statement:
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:
Input: head = [1,1,2]
Output: [1,2]
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.
Approach:
- Iterate through the linked list with a current pointer.
- For each node, compare its value with the next node.
- If they are equal, skip the next node using
curr.next = curr.next.next. - Otherwise, move to the next node.
Time and Space Complexity:
- Time Complexity: O(n)
- Space Complexity: O(1)
var deleteDuplicates = function(head) {
let curr = head;
while (curr && curr.next) {
if (curr.val === curr.next.val) {
curr.next = curr.next.next;
} else {
curr = curr.next;
}
}
return head;
};
struct ListNode* deleteDuplicates(struct ListNode* head) {
struct ListNode* curr = head;
while (curr && curr->next) {
if (curr->val == curr->next->val) {
struct ListNode* temp = curr->next;
curr->next = curr->next->next;
free(temp);
} else {
curr = curr->next;
}
}
return head;
}
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
ListNode* curr = head;
while (curr && curr->next) {
if (curr->val == curr->next->val) {
ListNode* temp = curr->next;
curr->next = curr->next->next;
delete temp;
} else {
curr = curr->next;
}
}
return head;
}
};
class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode curr = head;
while (curr != null && curr.next != null) {
if (curr.val == curr.next.val) {
curr.next = curr.next.next;
} else {
curr = curr.next;
}
}
return head;
}
}
class Solution(object):
def deleteDuplicates(self, head):
"""
:type head: Optional[ListNode]
:rtype: Optional[ListNode]
"""
curr = head
while curr and curr.next:
if curr.val == curr.next.val:
curr.next = curr.next.next
else:
curr = curr.next
return head
public class Solution {
public ListNode DeleteDuplicates(ListNode head) {
ListNode curr = head;
while (curr != null && curr.next != null) {
if (curr.val == curr.next.val) {
curr.next = curr.next.next;
} else {
curr = curr.next;
}
}
return head;
}
}
