[LeetCode] Linked List Components

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题目描述

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.

输入包括一个单链表以及一个数组，单链表中的节点不重复。数组存储图的所有节点，要求求出单链表中有多少个连通分量。
此题可以使用hash table存储图的所有节点，顺序遍历单链表，如果单链表连续几个节点都是图中的节点，则继续向前遍历查找。在找到第一不是图节点的时候，统计连通图数的变量加一。

C++ 实现

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/**
 * 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> g(G.begin(), G.end());
        int res = 0;
        while(head)
        {
            int component_size = 0;
            while (head && g.count(head->val))
            {
                component_size++;
                head = head->next;
            }
            
            if (component_size > 0)
            {
                res++;
            }
            
            if (head)
            {
                head = head->next;
            }
        }
        return res;
    }
};
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