Description

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.

Solution

HashSet, O(n), S(k)

n = list length
k = G.length
在list中找出现在G的连续节点组成的group数。很简单，遍历list的同时，查询当前节点是否出现在G中即可。

可以在new component start的时候increase count：

/**
 * 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> set = new HashSet<>();
        set.addAll(IntStream.of(G).boxed().collect(Collectors.toList()));
        boolean isNewComponentStart = true;
        int components = 0;
        
        while (head != null) {
            if (!set.contains(head.val)) {
                isNewComponentStart = true;
            } else {
                // increase count when new component started
                if (isNewComponentStart) {
                    ++components;
                }
                isNewComponentStart = false;
            }
            
            head = head.next;
        }
        
        return components;
    }
}

也可以在当前component end的时候increase count：

/**
* 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> set = new HashSet<>();
       set.addAll(IntStream.of(G).boxed().collect(Collectors.toList()));
       int components = 0;
       
       while (head != null) {
           // increase count when current component ends
           if (set.contains(head.val)
              && (head.next == null || !set.contains(head.next.val))) {
               ++components;
           }
           
           head = head.next;
       }
       
       return components;
   }
}