题目描述

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.
题意：给一个链表，返回中间节点，如果有两个中间节点，那么返回后面的那个。

样例

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.

算法1

题解：典型的快慢指针做法。慢指针一次走一个，快指针一次走两个，快指针走到链表尾部的时候，慢指针所在位置就是答案。

C++ 代码

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

如果题目要求返回靠前的那个，可以设置一个虚拟头节点，让快慢指针同时从虚拟节点出发。

    ListNode* middleNode(ListNode* head) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* slow = dummy;
        ListNode* quick = dummy;
        while(quick !=NULL && quick->next != NULL)
        {
            slow = slow->next;
            quick = quick->next->next;
        }
        return slow;
    }