leetcode 817. Linked list component (java)

Question description

Given the head node of the linked list, each node on the linked list has a unique integer value. At the same time, the list nums is given, which is a subset of the integer values ​​in the above linked list.
Returns the number of components in the list nums. The component is defined here as: the set of values ​​of the longest continuous node in the linked list (the value must be in the list nums).

Example 1:

Input: head = [0,1,2,3], nums = [0,1,3]
Output: 2
Explanation: In the linked list, 0 and 1 are connected, and nums does not contain 2, so [0, 1] is a component of nums. Similarly, [3] is also a component, so 2 is returned.

Example 2:

Input: head = [0,1,2,3,4], nums = [0,3,1,4]
Output: 2
Explanation: In the linked list, 0 and 1 are connected, 3 and 4 are are connected, so [0, 1] and [3, 4] are two components, so 2 is returned.

Tip:
The number of nodes in the linked list is n
1 <= n <= 10^4
0 <= Node.val < n
All values ​​in Node.val are different
1 <= nums.length <= n
0 <= nums[i] < n
All values ​​in nums are different

HashSet simulation

Just simulate according to the meaning of the question: In order to easily determine whether a certain node.val exists in nums, we first use the Set structure to dump all nums[i], and then check the number of consecutive segments (components) each time .

Code demo:

/**
 * 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) {
    
    
        Set<Integer>set = new HashSet<>();
        int count = 0;
        //方便进行判断，如果用数组判断，那么每次都要循环，这个优化很重要。
        for(int x : nums){
    
    
            set.add(x);
        }
        while(head != null){
    
    
          if(set.contains(head.val)){
    
    
            while(head != null && set.contains(head.val)){
    
    
                head = head.next;
            }
            count++;
          }else{
    
    
              head = head.next;
          }
        }
        return count;
    }
}