LeetCode – Remove Linked List Elements (Java)

Remove all elements from a linked list of integers that have value val.

Example

Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 --> 5

Java Solution

The key to solve this problem is using a helper node to track the head of the list.

public ListNode removeElements(ListNode head, int val) {
    ListNode helper = new ListNode(0);
    helper.next = head;
    ListNode p = helper;
 
    while(p.next != null){
        if(p.next.val == val){
            ListNode next = p.next;
            p.next = next.next; 
        }else{
            p = p.next;
        }
    }
 
    return helper.next;
}

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West

don’t be so dork add additional check on begining
Anshul

ListNode removeKFromList(ListNode l, int k) {

if(null == l) {
return null;
}

ListNode pointer = l;

while (pointer.next != null) {
if (pointer.next.value == k) pointer.next = pointer.next.next;
else pointer = pointer.next;
}

return l.value == k ? l.next : l;

}
Saurabh

Throws null pointer at if (pointer.next.val == val) pointer.next = pointer.next.next; for a LL of size 1
someone

void del_node (node_t **l, int value) {
node_t *prev = NULL;
node_t *cur = *l;

while (cur) {
if (cur->data == value) {
node_t *t = cur;
if (prev) { /* Not list head */
prev->next = cur->next;
} else { /* List head */
*l = cur->next;
}
cur = cur->next;
free(t);
} else {
prev = cur;
cur = cur->next;
}
}
}
Rakesh Venkatesh

No need to deference as it will be garbage collected
cabbar

leak*
cabbar

Do you dereference deleted node? Because if you don’t isn’t it memory lead?
Vimukthi Weerasiri

Without a helper node. Comments are welcome 🙂

public ListNode removeElements(ListNode head, int val) {
if (head == null) return null;
ListNode pointer = head;
while (pointer.next != null) {
if (pointer.next.val == val) pointer.next = pointer.next.next;
else pointer = pointer.next;
}
return head.val == val ? head.next : head;
}

LeetCode – Remove Linked List Elements (Java)

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