Remove Duplicates from Sorted List II LeetCode Solution | Easy Approach

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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.

Remove Duplicates from Sorted List II Solutions

Time: O(n)
Space: O(1)

C++

class Solution {
 public:
  ListNode* deleteDuplicates(ListNode* head) {
    ListNode dummy(0, head);
    ListNode* prev = &dummy;

    while (head) {
      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;
  }
};

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;
  }
}

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

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