Formatted question description: https://leetcode.ca/all/817.html

817. Linked List Components (Medium)

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 head1 <= 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.

Companies:
Google

Related Topics:
Linked List

Solution 1.

// OJ: https://leetcode.com/problems/linked-list-components/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int numComponents(ListNode* head, vector<int>& G) {
        unordered_set<int> s(G.begin(), G.end());
        int ans = 0;
        while (true) {
            while (head && s.find(head->val) == s.end()) head = head->next;
            if (!head) break;
            ++ans;
            while (head && s.find(head->val) != s.end()) head = head->next;
        }
        return ans;
    }
};

Java

  • /**
     * 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> subset = new HashSet<Integer>();
            for (int value : G)
                subset.add(value);
            boolean inComponent = false;
            int components = 0;
            ListNode node = head;
            while (node != null) {
                int value = node.val;
                if (subset.contains(value)) {
                    if (!inComponent) {
                        inComponent = true;
                        components++;
                    }
                } else {
                    if (inComponent)
                        inComponent = false;
                }
                node = node.next;
            }
            return components;
        }
    }
    
  • // OJ: https://leetcode.com/problems/linked-list-components/
    // Time: O(N)
    // Space: O(N)
    class Solution {
    public:
        int numComponents(ListNode* head, vector<int>& G) {
            unordered_set<int> s(G.begin(), G.end());
            int ans = 0;
            while (true) {
                while (head && s.find(head->val) == s.end()) head = head->next;
                if (!head) break;
                ++ans;
                while (head && s.find(head->val) != s.end()) head = head->next;
            }
            return ans;
        }
    };
    
  • # 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
            """
            groups = 0
            subset = set(G)
            while head:
                if head.val in subset and (not head.next or head.next.val not in subset):
                    groups += 1
                head = head.next
            return groups
    

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