leetcode-203-easy

Remove Linked List Elements

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

思路一: 递归，node 节点 == val, 递归调用 node.next

public ListNode removeElements(ListNode head, int val) {
    if (head == null) return head;

    if (head.val == val) {
        return removeElements(head.next, val);
    } else {
        head.next = removeElements(head.next, val);
        return head;
    }
}

思路二：迭代，创建一个头保存链表的引用，还有一个关键是用于遍历的 node，指向当前要判断 node

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