Remove Duplicates from Sorted List II LeetCode Solution

Problem – Remove Duplicates from Sorted List II LeetCode Solution

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 LeetCode Solution in Java

public ListNode deleteDuplicates(ListNode head) {
        if(head==null) return null;
        ListNode FakeHead=new ListNode(0);
        FakeHead.next=head;
        ListNode pre=FakeHead;
        ListNode cur=head;
        while(cur!=null){
            while(cur.next!=null&&cur.val==cur.next.val){
                cur=cur.next;
            }
            if(pre.next==cur){
                pre=pre.next;
            }
            else{
                pre.next=cur.next;
            }
            cur=cur.next;
        }
        return FakeHead.next;
    }

Remove Duplicates from Sorted List II LeetCode Solution in Python

def deleteDuplicates(self, head):
    dummy = pre = ListNode(0)
    dummy.next = head
    while head and head.next:
        if head.val == head.next.val:
            while head and head.next and head.val == head.next.val:
                head = head.next
            head = head.next
            pre.next = head
        else:
            pre = pre.next
            head = head.next
    return dummy.next

Remove Duplicates from Sorted List II LeetCode Solution in C++

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (!head) return 0;
        if (!head->next) return head;
        
        int val = head->val;
        ListNode* p = head->next;
        
        if (p->val != val) {
            head->next = deleteDuplicates(p);
            return head;
        } else {
            while (p && p->val == val) p = p->next;
            return deleteDuplicates(p);
        }
    }
};

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