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  • 817. Linked List Components

    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;
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10185134.html
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