Leetcode Remove Duplicates from Sorted List II problem solution

YASH PAL,

In this Leetcode Remove Duplicates from Sorted List II problem solution we have 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

Leetcode Remove Duplicates from Sorted List II problem solution

Problem solution in Python.

class Solution(object):
    def deleteDuplicates(self, head):
        if not head or not head.next :
            return head
        p = head
        while p.next and p.val == p.next.val:
            p = p.next
        if p != head:
            head = p.next
            return self.deleteDuplicates(head)
        head.next = self.deleteDuplicates(head.next)
        return head

Problem solution in Java.

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) return head;
        
        ListNode pre = new ListNode(-1, head);
        ListNode curr = head;
        ListNode dummyHead = pre;
        
        while (curr != null && curr.next != null) {
            if (curr.val == curr.next.val) {
                while (curr != null && curr.next != null && curr.val == curr.next.val) {
                    curr = curr.next;
                }
                pre.next = curr.next;
            } else {
                pre = curr;
            }
            curr = curr.next;
        }
        
        return dummyHead.next;
    }
}

Problem solution in C++.

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head==NULL) return head;
        ListNode *p0,*p1;
        p0 = new ListNode(0); /*It is convenient to create a node before head*/
        p0->next = head;
        p1 = p0;
        while(p1->next!=NULL){  /*remove the duplicates after and next to p1*/
            ListNode *tmp = p1->next;
            int current = tmp->val, count = 1;
            tmp = tmp->next;
            while(tmp != NULL && tmp->val == current){
                count++;
                tmp = tmp->next;
            }
            if(count > 1) p1->next = tmp;
            else p1 = p1->next;
        }
        return p0->next;
    }
};

Problem solution in C.

if(head == NULL)
    return NULL;
    
struct ListNode *p1,*p2;
p1 = head ; p2 = head;

while(p2 != NULL && p2->val == p1->val)
    p2 = p2->next;

if(p1->next == p2)
{
    p1->next = deleteDuplicates(p2);
    return p1;
}
else
    return deleteDuplicates(p2);

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