82. Remove Duplicates from Sorted List II LeetCode Solution
In this guide, you will get 82. Remove Duplicates from Sorted List II LeetCode Solution with the best time and space complexity. The solution to Remove Duplicates from Sorted List II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Remove Duplicates from Sorted List II
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.
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
82. Remove Duplicates from Sorted List II LeetCode Solution in C++
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
ListNode dummy(0, head);
ListNode* prev = &dummy;
while (head != nullptr) {
while (head->next && head->val == head->next->val)
head = head->next;
if (prev->next == head)
prev = prev->next;
else
prev->next = head->next;
head = head->next;
}
return dummy.next;
}
}; /* code provided by PROGIEZ */
82. Remove Duplicates from Sorted List II LeetCode Solution in Java
class Solution {
public ListNode deleteDuplicates(ListNode head) {
ListNode dummy = new ListNode(0, head);
ListNode prev = dummy;
while (head != null) {
while (head.next != null && head.val == head.next.val)
head = head.next;
if (prev.next == head)
prev = prev.next;
else
prev.next = head.next;
head = head.next;
}
return dummy.next;
}
} // code provided by PROGIEZ
82. Remove Duplicates from Sorted List II LeetCode Solution in Python
class Solution:
def deleteDuplicates(self, head: ListNode) -> ListNode:
dummy = ListNode(0, head)
prev = dummy
while head:
while head.next and head.val == head.next.val:
head = head.next
if prev.next == head:
prev = prev.next
else:
prev.next = head.next
head = head.next
return dummy.next # code by PROGIEZ
