1 """ 2 We are given head, the head node of a linked list containing unique integer values. 3 4 We are also given the list G, a subset of the values in the linked list. 5 6 Return the number of connected components in G, where two values are connected if they appear consecutively in the linked list. 7 8 Example 1: 9 10 Input: 11 head: 0->1->2->3 12 G = [0, 1, 3] 13 Output: 2 14 Explanation: 15 0 and 1 are connected, so [0, 1] and [3] are the two connected components. 16 17 Example 2: 18 19 Input: 20 head: 0->1->2->3->4 21 G = [0, 3, 1, 4] 22 Output: 2 23 Explanation: 24 0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components. 25 26 """ 27 class Solution: 28 def numComponents(self, head, G): 29 set_G = set(G) #试了用list也能通过，set是为了规范数据集 30 p = head 31 count =0 32 while(p): 33 if p.val in set_G and (p.next == None or p.next!=None and p.next.val not in set_G): 34 #！！！if条件句关键 35 #将G中的元素放入set 或者字典中用于查找。 36 # 遍历链表， 链表中的元素被计数的条件是， 37 # 如果当前元素在G中且下一个元素不在G中（或者为None）, 38 # 那么当前的元素属于一个component。 39 count += 1 40 p = p.next 41 return count