Remove Linked List Elements


Problem Statement :


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.

 

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:

The number of nodes in the list is in the range [0, 104].
1 <= Node.val <= 50
0 <= val <= 50


Solution :



title-img 


                            Solution in C :

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeElements(struct ListNode* head, int val) {
    struct ListNode *temp = (struct ListNode*)malloc(sizeof(struct ListNode));
    temp->next = head;
    struct ListNode *curr = temp;
    while(curr->next != NULL ){
        if(curr->next->val == val) curr->next = curr->next->next;
        else curr = curr->next;
    }
    return temp->next;
}
                        


                        Solution in 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* removeElements(ListNode* head, int val) {
        ListNode *temp = new ListNode(0);
        temp->next = head;
        ListNode *curr = temp;
        while(curr->next != NULL ){
            if(curr->next->val == val) curr->next = curr->next->next;
            else curr = curr->next;
        }
        return temp->next;
    }
};
                    


                        Solution in 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 removeElements(ListNode head, int val) {
        ListNode temp = new ListNode(0) , curr = temp;
        temp.next = head;
        while(curr.next != null ){
            if(curr.next.val == val) curr.next = curr.next.next;
            else curr = curr.next;
        }
        return temp.next;
    }
}
                    


                        Solution in Python : 
                            
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        temp = ListNode(0)
        temp.next = head
        prev, curr = temp, head
        while curr:
            if curr.val == val:prev.next = curr.next
            else:prev = curr
            curr = curr.next
        return temp.next


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