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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0817_linked_list_components;

// #Medium #Hash_Table #Linked_List #2022_03_23_Time_7_ms_(76.10%)_Space_54_MB_(54.99%)

import com_github_leetcode.ListNode;
import java.util.HashSet;

/*
 * 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; }
 * }
 */
/**
 * 817 - Linked List Components\.
 *
 * Medium
 *
 * 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/07/22/lc-linkedlistcom1.jpg)
 *
 * **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:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/07/22/lc-linkedlistcom2.jpg)
 *
 * **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 <= 10⁴
 * *   `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**.
**/
public class Solution {
    public int numComponents(ListNode head, int[] nums) {
        HashSet 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;
    }
}