Description#
You are given the head
of a linked list containing unique integer values and an integer array nums
that is a subset of the linked list values.
Return the number of connected components in nums
where two values are connected if they appear consecutively in the linked list.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/0817.Linked%20List%20Components/images/lc-linkedlistcom1.jpg)
Input: head = [0,1,2,3], nums = [0,1,3]
Output: 2
Explanation: 0 and 1 are connected, so [0, 1] and [3] are the two connected components.
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/0817.Linked%20List%20Components/images/lc-linkedlistcom2.jpg)
Input: head = [0,1,2,3,4], nums = [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.
Constraints:
- The number of nodes in the linked list is
n
. 1 <= n <= 104
0 <= Node.val < n
- All the values
Node.val
are unique. 1 <= nums.length <= n
0 <= nums[i] < n
- All the values of
nums
are unique.
Solutions#
Solution 1#
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| # 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: Optional[ListNode], nums: List[int]) -> int:
ans = 0
s = set(nums)
while head:
while head and head.val not in s:
head = head.next
ans += head is not None
while head and head.val in s:
head = head.next
return ans
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| /**
* 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) {
int ans = 0;
Set<Integer> s = new HashSet<>();
for (int v : nums) {
s.add(v);
}
while (head != null) {
while (head != null && !s.contains(head.val)) {
head = head.next;
}
ans += head != null ? 1 : 0;
while (head != null && s.contains(head.val)) {
head = head.next;
}
}
return ans;
}
}
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| /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
int numComponents(ListNode* head, vector<int>& nums) {
unordered_set<int> s(nums.begin(), nums.end());
int ans = 0;
while (head) {
while (head && !s.count(head->val)) head = head->next;
ans += head != nullptr;
while (head && s.count(head->val)) head = head->next;
}
return ans;
}
};
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| /**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func numComponents(head *ListNode, nums []int) int {
s := map[int]bool{}
for _, v := range nums {
s[v] = true
}
ans := 0
for head != nil {
for head != nil && !s[head.Val] {
head = head.Next
}
if head != nil {
ans++
}
for head != nil && s[head.Val] {
head = head.Next
}
}
return ans
}
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| /**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function numComponents(head: ListNode | null, nums: number[]): number {
const set = new Set<number>(nums);
let res = 0;
let cur = head;
let inSet = false;
while (cur != null) {
if (set.has(cur.val)) {
if (!inSet) {
inSet = true;
res++;
}
} else {
inSet = false;
}
cur = cur.next;
}
return res;
}
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| // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
// pub val: i32,
// pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
// #[inline]
// fn new(val: i32) -> Self {
// ListNode {
// next: None,
// val
// }
// }
// }
use std::collections::HashSet;
impl Solution {
pub fn num_components(head: Option<Box<ListNode>>, nums: Vec<i32>) -> i32 {
let set = nums.into_iter().collect::<HashSet<i32>>();
let mut res = 0;
let mut in_set = false;
let mut cur = &head;
while let Some(node) = cur {
if set.contains(&node.val) {
if !in_set {
in_set = true;
res += 1;
}
} else {
in_set = false;
}
cur = &node.next;
}
res
}
}
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| /**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number[]} nums
* @return {number}
*/
var numComponents = function (head, nums) {
const s = new Set(nums);
let ans = 0;
while (head) {
while (head && !s.has(head.val)) {
head = head.next;
}
ans += head != null;
while (head && s.has(head.val)) {
head = head.next;
}
}
return ans;
};
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