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 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.

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;
    }
}

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