[Leetcode] 817. Linked List Components 解題報告

題目：

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.

思路：

我們順次檢查連結清單中的節點是否出現在G中，如果出現了，則增加一個統計，并連續跟蹤直到目前的連續序列結束。這樣周遊完成一遍清單，就可以獲得預期答案。

為了加速查找，我們将G中的元素放入哈希表中，這樣可以将算法的時間複雜度降低到O(n)，其中n是連結清單的長度。算法的空間複雜度則為O(m)，其中m是G中元素的個數。由題意易知m < n。

代碼：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    int numComponents(ListNode* head, vector<int>& G) {
        unordered_set<int> hash;
        for (int &g : G) {
            hash.insert(g);
        }
        ListNode *node = head;
        int ans = 0;
        while (node != NULL) {
            if (hash.count(node->val) > 0) {
                ++ans;
                while (node != NULL && hash.count(node->val) > 0) {
                    node = node->next;
                }
            }
            if (node != NULL) {
                node = node->next;
            }
        }
        return ans;
    }
};