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

Example 2:

```Input:
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.

## Rust Solution

``````struct Solution;
use rustgym_util::*;
use std::collections::HashSet;
use std::iter::FromIterator;

impl Solution {
let hs: HashSet<i32> = HashSet::from_iter(g);
let mut open = false;
let mut res = 0;
while let Some(node) = p {
if hs.contains(&node.val) {
if !open {
open = true;
}
} else {
if open {
open = false;
res += 1;
}
}
p = node.next;
}
if open {
res += 1;
}
res
}
}

#[test]
fn test() {
let head = list!(0, 1, 2, 3);
let g = vec![0, 1, 3];
let res = 2;