Given the head of a singly linked list, return the middle node of the linked list.
If there are two middle nodes, return the second middle node.
Example 1:
Input: head = [1,2,3,4,5]
Output: [3,4,5]
Explanation: The middle node of the list is node 3.
Example 2:
Input: head = [1,2,3,4,5,6]
Output: [4,5,6]
Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.
Constraints:
The number of nodes in the list is in the range [1, 100].
1 <= Node.val <= 100
// 我的代码
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode() : val(0), next(nullptr) {}* ListNode(int x) : val(x), next(nullptr) {}* ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* middleNode(ListNode* head) {ListNode * slow = head;ListNode * fast = head;while(fast != NULL && fast->next != NULL){fast = fast->next->next;slow = slow->next;//cout<val<
注意while循环的条件,fast->next也要非NULL才OK。
快慢指针除了做循环检测,还可以做中点检测。