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;
}

If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:

<pre><code> 
String foo = "bar";
</code></pre>

N G Sundarasekaran

public static ListNode removeNodes(ListNode listHead, int x) {
ListNode head = listHead;
while(head != null && head.data == x) {
head = head.next;
}

if(head != null) {
ListNode node = head;
ListNode nextNode = node.next;
while(nextNode != null) {
if(nextNode.data > x) {
node.next = nextNode.next;
} else {
node = nextNode;
}
nextNode = node.next;
}
}

return head;

}
Alik Elzin

–
A solution without helper below. It has more edge cases like updating root.

Node remove(Node root, int val) { Node prev = null;
for (Node n = root; n != null ; n = n.next) { if (n.val != val) { if (prev == null) { root == n; } prev = n; continue; } if (prev != null) { prev.next = n.next; } } if (prev == null) return null;
return root; }
Hemant Kumar Kushwaha

How to pass the ListNode head parameter while calling this function in main method.
sabari balaji

thanks for this post this is informative
java training in velachery
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)

Related posts: