Deletion in mid in linked lists

#1

sir, i tried this question lnk:https://leetcode.com/problems/remove-linked-list-elements/
but it is giving undefined memory i think some problem related with function
/**

  • 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 remove(ListNode *temp,ListNode *prev)
    {
    ListNode *del = temp;
    prev -> next = temp -> next;
    temp = temp -> next;
    delete(del);
    cout<val<<" "<val<<endl;
    }

    ListNode* removeElements(ListNode* head, int data)
    {
    ListNode *temp = head;
    ListNode *prev = NULL;
    while(temp->next!=NULL)
    {
    if(temp->val == data)
    {
    remove(temp,prev);
    cout<val<<" "<val<<endl;

         }
     else
     {
         prev = temp;
         temp = temp->next;
     }
 }
 
 return head;

    }
    };

#2

sir use this link to open the code instead link:https://ide.codingblocks.com/s/291784

#3

hello @khushi91200

image
image727×158 4.42 KB

a)pass them as refernce
b) prev can be NULL so handle that case separately.

c)
image
here it should be temp!=NULL;
d) ur head can also be deleted to add such cases as well

#4

how to pass it as reference sir?

#5

sir i know the rest case i will do it i am confused only how to do it by refernce coz pointers are passed by reference

#6

like this->
void func( node * &p){

}

#7

thanks sir it worked but sir pointers are passed by reference

#8

so why we are explicitly making it as reference

#9

pointers are also a variable .they just contain address.
so whathever behaviour we expect from any variable of any other type.
same holds true for pointers.

#10

means pointers are passed by value not reference?

#11

yeah…
they r special because they contain addresses .

#12

ok sir thankyou!!!