876. Middle of the Linked List

 876. Middle of the Linked List ( https://leetcode.com/problems/middle-of-the-linked-list/ )

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

思路：
使用快慢雙指標
快的一次走兩步
慢的一次走一步
快的走到底時，慢的就是走到中間

slow 跟 fast 都從 1 開始走
若 fast 能夠前進兩步，那 slow 就前進一步
若 fast 只能前進一步，那 slow 停在原地

/**

 * 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 = fast->next;

            if (fast != NULL)

            {

                fast = fast->next;

                slow = slow->next;

            }

        }

        return slow;

    }

};

參考：
https://www.youtube.com/watch?v=f4xocU8WIaU