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.
Note:
- If
N
is the length of the linked list given byhead
,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.
c++代码
/** * 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) { int res=0; set<int> s; for(auto&g: G) s.insert(g); bool enter=false; //bool leave=false; while(head){ if(s.find(head->val)!=s.end()) { if(enter==false) { enter=true; res+=1; } }else { enter=false; } head=head->next; } return res; } };
python代码
# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def numComponents(self, head, G): """ :type head: ListNode :type G: List[int] :rtype: int """ cnt=0 enter=False while head is not None: if head.val in G: if not enter: enter=True cnt+=1 else: enter=False #print "head.val: {}, cnt: {}".format(head.val,cnt) head=head.next return cnt