# 【LeetCode】 876. Middle of the Linked List

## Description
> Given a non-empty, singly linked list with head node head, return a middle node of linked list.
> If there are two middle nodes, return the second middle node.
> Note:
> The number of nodes in the given list will be between 1 and 100.

> 給一個非空的、單獨的linked list的開頭head，回傳最中間的節點。
> 如果有兩個中間點，回傳第二個點。
> 注意：
> 點的數量只會介於1到100之間。

## Example:

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3.  (The judge's serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.


Example 2:

Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

## Solution

* 這題很基礎也蠻常考的，強烈建議可以好好記起來。
* 如果先算list的長度再重跑，要跑1.5l，這樣實在太慢了。
* 最好的方法是有兩個點一起跑，第一個一次跑一步，第二個一次跑兩步。
* 當第二個點看到尾巴時，第一個點就是中間點。
* 特別注意點的數量為單數和雙數時要分別處理。

### Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        ListNode* one = head;
        ListNode* two = head;
        
        while(1)
        {
            if(two->next == NULL)
                return one;
            if(two->next->next == NULL)
                return one->next;
            
            one = one->next;
            two = two->next->next;
        }
    }
};

###### tags: `LeetCode` `C++`