817. Linked List Components

Problem

You are given the head of a linked list containing unique integer values and an integer array nums that is a subset of the linked list values.

Return **the number of connected components in *nums* where two values are connected if they appear consecutively in the linked list**.

  Example 1:

Input: head = [0,1,2,3], nums = [0,1,3]
Output: 2
Explanation: 0 and 1 are connected, so [0, 1] and [3] are the two connected components.

Example 2:

Input: head = [0,1,2,3,4], nums = [0,3,1,4]
Output: 2
Explanation: 0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components.

  Constraints:

• The number of nodes in the linked list is n.

• 1 <= n <= 10^4

• 0 <= Node.val < n

• All the values Node.val are unique.

• 1 <= nums.length <= n

• 0 <= nums[i] < n

• All the values of nums are unique.

Solution (Java)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int numComponents(ListNode head, int[] nums) {
        HashSet<Integer> set = new HashSet<>();
        for (int i : nums) {
            set.add(i);
        }
        int result = 0;
        while (head != null) {
            if (set.contains(head.val)) {
                while (head != null && set.contains(head.val)) {
                    head = head.next;
                }
                result++;
            } else {
                head = head.next;
            }
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).