README_EN.md

817. Linked List Components

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Description

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.

Solutions

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def numComponents(self, head: ListNode, nums: List[int]) -> int:
        s = set(nums)
        res, pre = 0, True
        while head:
            if head.val in s:
                if pre:
                    res += 1
                    pre = False
            else:
                pre = True
            head = head.next
        return res

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int numComponents(ListNode head, int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int num : nums) {
            s.add(num);
        }
        int res = 0;
        boolean pre = true;
        while (head != null) {
            if (s.contains(head.val)) {
                if (pre) {
                    ++res;
                    pre = false;
                }
            } else {
                pre = true;
            }
            head = head.next;
        }
        return res;
    }
}

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