817.md

Linked List Components

Description

link

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Solution

• See Code

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Code

O(n)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def numComponents(self, head: ListNode, G: List[int]) -> int:
        res = 0
        G = set(G)
        while head:
            if head.val in G:
                res += 1
                while head and head.val in G:
                    head = head.next
            else:
                head = head.next
        return res