leetcode817 Linked List Components

 """
 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.

 """
 class Solution:
     def numComponents(self, head, G):
         set_G = set(G)  #试了用list也能通过，set是为了规范数据集
         p = head
         count =0
         while(p):
             if p.val in set_G and (p.next == None or p.next!=None and p.next.val not in set_G):
                 #！！！if条件句关键
                 #将G中的元素放入set 或者字典中用于查找。
                 # 遍历链表， 链表中的元素被计数的条件是，
                 # 如果当前元素在G中且下一个元素不在G中（或者为None）,
                 # 那么当前的元素属于一个component。
                 count += 1
             p = p.next
         return count