817. Linked List Components.md

817. Linked List Components

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.

Solution

• Java

  • mine

    Runtime: 156 ms, faster than 5.08%, Memory Usage: 49.3 MB, less than 7.69% of Java online submissions

    public int numComponents(ListNode head, int[] G) {
        if (G.length == 1) {
            return 1;
        }
        int count = 0;
        int t = 0;
        while (head != null) {
            boolean c = contain(G, head.val);
            if (c) {
                t++;
            } else {
                count += (t == 0 ? 0 : 1);
                t = 0;
            }
            head = head.next;
        }
        if (t != 0) {
            count++;
        }
        return count;
    }
    
    public boolean contain(int[] G, int v) {
        for (int t : G) {
            if (t == v) {
                return true;
            }
        }
        return false;
    }
    

  • the most votes

    Runtime: 6 ms, faster than 49.38%,Memory Usage: 40.5 MB, less than 76.92% of Java online submissions

    public int numComponents(ListNode head, int[] G) {
        Set<Integer> setG = new HashSet<>();
        for (int i: G) setG.add(i);
        int res = 0;
        while (head != null) {
            if (setG.contains(head.val) && (head.next == null || !setG.contains(head.next.val))) res++;
            head = head.next;
        }
        return res;
    }