203. Remove Linked List Elements

recursion

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

naive方法，先移除头部，再移除中间。

public ListNode removeElements(ListNode head, int val) {
    // remove head
    while (head != null && head.val == val) head = head.next;
    // remove internal nodes
    ListNode node = head;
    while (node != null && node.next != null) {
        if (node.next != null && node.next.val == val) {
            node.next = node.next.next;
        } else {
            node = node.next;
        }
    }
    return head;
}

dummy node方法

public ListNode removeElements(ListNode head, int val) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode cur = dummy;
    while (head != null) {
        if (head.val != val) {
            cur.next = head;
            cur = head;
        }
        head = head.next;
    }
    cur.next = null; // in case that all elements in the list are same
    return dummy.next;
}