LeetCode 30 days challenge: Middle of the Linked List

題目名稱 Middle of the Linked List

題目描述：
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.

例子：

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.

我的理解
最初的想法是兩次遍歷，第一次遍歷可以得到鏈表的長度，就可以知道長度一半上取整的值是多少，故第二次遍歷可以找到中間位置。時間複雜度是o(kn)。

其實鏈表題很多情況下都會使用快慢指針。快慢指針有一個特點，就是快指針每次走兩步，慢指針每次走一步，相當於快指針比慢指針的速度大1。如果次數爲t，快指針就比慢指針多走t步, 好巧不巧慢指針也剛好走了t步。也就是說不管次數爲多少，慢指針剛剛好在快指針和起點的中間位置上。

C++ solution

class Solution {
public:
   ListNode* middleNode(ListNode* head) {
		ListNode *slow = head, *fast = head;
		while (fast && fast->next) {
			slow = slow->next;
			fast = fast->next->next;
		}
		return slow;
    }
};