题目:
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 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.
思路:
我们顺次检查链表中的节点是否出现在G中,如果出现了,则增加一个统计,并连续跟踪直到当前的连续序列结束。这样遍历完成一遍列表,就可以获得预期答案。
为了加速查找,我们将G中的元素放入哈希表中,这样可以将算法的时间复杂度降低到O(n),其中n是链表的长度。算法的空间复杂度则为O(m),其中m是G中元素的个数。由题意易知m < n。
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) { 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; } };