876. Middle of the Linked List

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.
找到中间结点，并返回。
1. 第一种方法是用计数器遍历链表得到链表长度，然后除二取整，再用循环直接迭加到中间结点。

struct ListNode* middleNode(struct ListNode* head) {
    if(!head)   return NULL;
    struct ListNode *current=head;
    int count=0;
    while(current){
        current=current->next;
        count++;
    }
    count/=2;
    current=head;//重新赋值多次，不然current是NULL报错
    while(count){
        current=current->next;
        count--;
    }
    return current; 
}

2. 第二种方法是用两个指针，快的走两步，慢的走一步，当快的走完的时候，慢的刚好停在中间。

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