817. Linked List Components - Medium

最新推荐文章于 2023-06-07 20:02:31 发布

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最新推荐文章于 2023-06-07 20:02:31 发布

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原文链接：http://www.cnblogs.com/fatttcat/p/10185134.html

本文介绍了一种算法，用于计算链表中特定值组成的连续组件数量。通过将给定值集合转换为哈希集并遍历链表，判断每个节点是否构成独立组件，实现了高效的时间复杂度O(n)和空间复杂度O(k)，其中k为给定值集合的长度。

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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.

先把G中的元素加入set

遍历linked list，如果当前指针指向节点的值在set中存在，并且下一个为空/下一个不存在，说明是一个独立的component，component数+1；

如果下一个不为空并且下一个值在set中存在，说明这仍然是同一个component，指针指向下一个元素

time: O(n), space: O(k)　　-- k: length of G

/**
 * 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> set = new HashSet<>();
        for(int g : G) {
            set.add(g);
        }
        
        int res = 0;
        while(head != null) {
            if(set.contains(head.val) && (head.next == null || !set.contains(head.next.val))) {
                res++;
            }
            head = head.next;
        }
        return res;
    }
}