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->3G = [0, 1, 3]Output: 2Explanation: 0 and 1 are connected, so [0, 1] and [3] are the two connected components.

Example 2:

Input: head: 0->1->2->3->4G = [0, 3, 1, 4]Output: 2Explanation: 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.

这道题让我们判断链表里面有几个节点是“connected components”，单个的节点在数组中出现算一个“connected components”，相连的几个节点在数组中出现，只算一个“connected components”。

我们可以用二分法查找当前节点是否在数组中出现，如果当前节点的后继节点同样存在，则将指针后移，如果不存在则res++。这里要加一个flag，判断之前的节点是否存在数组中。

解法一：

/** * 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
    
     & G) {        sort(G.begin(),G.end());        int res=0,flag=0;        while(head!=NULL){            while(head&&find(G,head->val)){                head=head->next;                flag=1;            }            if(flag) res++;            flag=0;            if(head&&head->next) head=head->next;            else return res;        }        return res;    }    bool find(vector
     
      & nums, int target) {        int left = 0, right = nums.size();        while (left < right) {            int mid = left + (right - left) / 2;            if (nums[mid] == target) return true;            else if (nums[mid] < target) left = mid + 1;            else right = mid;        }        return false;    }};
     
    

STL提供了一个容器可以使用count()直接查找，省去了写二分法，参考代码如下。

解法二

/** * 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
    
     & G) {        unordered_set
     
       setG (G.begin(), G.end());        int res = 0;        while (head != NULL) {            if (setG.count(head->val) && (head->next == NULL || !setG.count(head->next->val))) res++;            head = head->next;        }        return res;    }};