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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 head1 <= 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
        }
    }
    
    

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