链表的中间结点，简单的快慢指针问题

 

 

 

总结

 

struct ListNode* middleNode(struct ListNode* head) {
    struct ListNode*fast=head;
    struct ListNode*slow=head;

    while( fast && fast->next)//结束条件
    {
       slow=slow->next;
       fast=fast->next->next;
    }
    return slow;
}