Leetcode: Linked List Components

Linked List Components



Similar Problems:


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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## Blog link: https://code.dennyzhang.com/linked-list-components
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def numComponents(self, head, G):
        """
        :type head: ListNode
        :type G: List[int]
        :rtype: int
        """
        ## Basic Ideas: One pass for linked list
        ##              set
        ## Complexity: Time O(n), Space O(n)
        nodes_set = set(G)
        res = 0
        node = head
        new_start = True
        while node:
            if node.val in nodes_set:
                if new_start:
                    res += 1
                    new_start = False
            else:
                new_start = True
            node = node.next
        return res
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