« back

Remove Linked List Elements

Tags : linkedlist, recursion, leetcode, cpp, easy

Problem Statement - link #

Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

Examples #

Example 1:

Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]

Example 2:

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

Example 3:

Input: head = [7,7,7,7], val = 7
Output: []

Constraints #

Solutions #

/**
 * 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:
    void re(ListNode* head,int val){
        if(!head->next) return;
        if(head->next->val == val){
            head->next = head->next->next;
            re(head,val);
        }    
        else
            re(head->next,val);
        
    }
    ListNode* removeElements(ListNode* head, int val) {
        if(!val || !head) return head;
        ListNode* ptr = new ListNode(0,head);
        re(ptr,val);
        return ptr->next;
    }
};