We are given head, the head node of a linked list containing unique integer values.

We are also given the list G, a subset of the values in the linked list.

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

Example 1:

Input: 
head: 0->1->2->3
G = [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
G = [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.

Note:

• If N is the length of the linked list given by head, 1 <= N <= 10000.
• The value of each node in the linked list will be in the range [0, N - 1].
• 1 <= G.length <= 10000.
• G is a subset of all values in the linked list.

思路：loop一遍，然后用flag来标记之前是否找到，然后count一下即可。这题没有什么意义。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int numComponents(ListNode head, int[] G) {
        Set<Integer> hashset = new HashSet<Integer>();
        for(Integer i: G){
            hashset.add(i);
        }
        int count = 0;
        ListNode cur =head;
        boolean found = false;
        while(cur != null) {
            if(hashset.contains(cur.val)){
                if(!found){
                    found = true;
                    count++;
                }else {
                    cur = cur.next;
                }
            } else { 
                found = false;
                cur = cur.next;
            }
        }
        return count;
    }
}