83. Remove Duplicates from Sorted List

Description of Problem

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 1:

Input: head = [1,1,2]
Output: [1,2]

Example 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.

Solution

Tags: LinkedList

Code (Java)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null){
            return head;
        }

        ListNode node1 = head;
        ListNode node2 = head.next;
        while(node2 != null){
            if(node1.val == node2.val){
                node1.next = node2.next;
                node2 = node2.next;
            }
            else{
                node1 = node1.next;
                node2 = node2.next;
            }
        }

        return head;
        
    }
}

Code (C++)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if (head == nullptr || head->next == nullptr){
            return head;
        }

        ListNode* node1 = head;
        ListNode* node2 = head->next;
        while(node2 != nullptr){
            if(node1->val == node2->val){
                node1->next = node2->next;
                node2 = node2->next;
            }
            else{
                node1 = node1->next;
                node2 = node2->next;
            }
        }

        return head;
    }
};

Complexity

n is the number of elements in the linked list

Time Complexity

\(T(n) = O(n)\)

Auxiliary Space

\(S(n) = O(1)\)

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