[LeetCode] Linked List Components 链表组件

时间:2023-12-09 21:08:07

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 head1 <= 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.

这道题给了我们一个链表,又给了我们一个结点值数组,里面不一定包括了链表中所有的结点值。让我们返回结点值数组中有多少个相连的组件,因为缺失的结点值会将原链表断开,实际上就是让我们求有多少个相连的子链表,题目中给的例子很好的说明题意。这道题并不需要什么特别高深的技巧,难懂的算法,直接按题目的要求来找就可以了。首先,为了快速的在结点值数组中查找某个结点值是否存在,我们可以将所有的结点值放到一个HashSet中,这样我们就能在常数级的时间复杂度中查找。然后我们就可以来遍历链表了,对于遍历到的每个结点值,我们只有两种情况,在或者不在HashSet中。不在HashSet中的情况比较好办,说明此时断开了,而在HashSet中的结点,有可能是该连续子链表的起始点,或者是中间的某个点,而我们的计数器对该子链表只能自增1,所以我们需要想办法来hanlde这种情况。博主最先想到的办法是先处理不在HashSet中的结点,处理方法就是直接跳到下一个结点。那么对于在HashSet中的结点,我们首先将计数器res自增1,然后再来个循环,将之后所有在集合中的结点都遍历完,这样才不会对同一个子链表多次增加计数器,参见代码如下:

解法一:

class Solution {
public:
int numComponents(ListNode* head, vector<int>& G) {
int res = ;
unordered_set<int> nodeSet(G.begin(), G.end());
while (head) {
if (!nodeSet.count(head->val)) {
head = head->next;
continue;
}
++res;
while (head && nodeSet.count(head->val)) {
head = head->next;
}
}
return res;
}
};

我们可以稍稍修改代码,使其更加简洁,我们在遍历的时候进行判断,如果当前结点在集合中,并且当前结点是尾结点或者下一个结点不在集合中的时候,我们让计数器自增1,通过这种操作,我们不会多加也不会漏加计数器,参见代码如下:

解法二:

class Solution {
public:
int numComponents(ListNode* head, vector<int>& G) {
int res = ;
unordered_set<int> nodeSet(G.begin(), G.end());
while (head) {
if (nodeSet.count(head->val) && (!head->next || !nodeSet.count(head->next->val))) {
++res;
}
head = head->next;
}
return res;
}
};

参考资料:

https://leetcode.com/problems/linked-list-components/description/

https://leetcode.com/problems/linked-list-components/solution/

https://leetcode.com/problems/linked-list-components/discuss/123842/C++JavaPython-Easy-and-Concise-Solution-with-Explanation

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