# 1. Question

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:
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Input:
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G = [0, 1, 3]
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Output: 2
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Explanation:
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0 and 1 are connected, so [0, 1] and  are the two connected components.
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Example 2:
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Input:
2
3
G = [0, 3, 1, 4]
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Output: 2
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Explanation:
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0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components.
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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.

# 2. Implementation

(1) HashSet
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/**
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public int numComponents(ListNode head, int[] G) {
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Set<Integer> set = new HashSet();
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for (int node : G) {
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}
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int res = 0;
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++res;
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}
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}
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return res;
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}
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}
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# 3. Time & Space Complexity

HashSet: 时间复杂度O(n), 空间复杂度O(n)