Formatted question description: https://leetcode.ca/all/817.html

817. Linked List Components (Medium)

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.

Companies:
Google

Related Topics:
Linked List

Solution 1.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int numComponents(ListNode head, int[] G) {
        Set<Integer> subset = new HashSet<Integer>();
        for (int value : G)
            subset.add(value);
        boolean inComponent = false;
        int components = 0;
        ListNode node = head;
        while (node != null) {
            int value = node.val;
            if (subset.contains(value)) {
                if (!inComponent) {
                    inComponent = true;
                    components++;
                }
            } else {
                if (inComponent)
                    inComponent = false;
            }
            node = node.next;
        }
        return components;
    }
}

############

/**
 * 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;
    }
}

// OJ: https://leetcode.com/problems/linked-list-components/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int numComponents(ListNode* head, vector<int>& G) {
        unordered_set<int> s(G.begin(), G.end());
        int ans = 0;
        while (true) {
            while (head && s.find(head->val) == s.end()) head = head->next;
            if (!head) break;
            ++ans;
            while (head && s.find(head->val) != s.end()) head = head->next;
        }
        return ans;
    }
};

# 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

############

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def numComponents(self, head, G):
        """
        :type head: ListNode
        :type G: List[int]
        :rtype: int
        """
        groups = 0
        subset = set(G)
        while head:
            if head.val in subset and (not head.next or head.next.val not in subset):
                groups += 1
            head = head.next
        return groups

/**
 * 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
}

/**
 * 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;
}

/**
 * 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;
};

// 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
    }
}

817 - Linked List Components

817. Linked List Components (Medium)

Solution 1.

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817. Linked List Components (Medium)

Solution 1.

All Problems

All Solutions