876. Middle of the Linked List。

版权声明：本文为博主原创文章，转载请标明出处：http://blog.csdn.net/leafage_m https://blog.csdn.net/Leafage_M/article/details/85718175

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.

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.

Note:

The number of nodes in the given list will be between 1 and 100.

原文链接：https://leetcode.com/problems/middle-of-the-linked-list/

---

找到链表的中间位置，从中间位置返回一个链表。

---

简单题，快慢指针。

class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        if(head==NULL || head->next==NULL)
            return head;

        ListNode* slow = head;
        ListNode* fast = head;
        while( fast!=NULL && fast->next!=NULL) {
            slow = slow->next;
            fast = fast->next->next;
            //cout << slow->val << " ";
        }
        return slow;
    }
};