# Leetcode 解題速記 876. Middle of the Linked List
###### tags: `LeetCode` `C++`
題敘:
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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.



測資範圍:
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* The number of nodes in the list is in the range [1, 100].
* 1 <= Node.val <= 100

解題筆記:
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本題使用C的struct實作linked list，以及龜兔賽跑演算法(雙指針)
`struct ListNode *next` 儲存的是linked list中下一筆資料的Address
因此只要開頭將slow、fast兩個指針初始化成第一個Address(`*head`)
**使用->運算子(等同於`(*fast).next`)**
fast一次前進兩個Address、反之slow一次只前進一個Address
當fast走到linked list末位，即NULL時終止迴圈
此時slow會剛好是linked list中的middleNode
return slow即可

程式碼:
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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* middleNode(struct ListNode* head){
    struct ListNode* fast=head;
    struct ListNode* slow=head;
    
    while(fast!=NULL && fast->next!=NULL){
        fast=fast->next->next;
        slow=slow->next;
    }
    return slow;
}