链表的中间结点

点击链接做题

思路代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
typedef struct ListNode ListNode;
struct ListNode* middleNode(struct ListNode* head) {
    ListNode* slow = head, *fast = head;
    //慢指针每次走1步，快指针每次走2步
    while(fast && fast->next){//不能写while(fast->next && fast)
        slow = slow->next;
        fast = fast->next->next;
    }
    //此时slow指向的结点刚好就是中间结点
    return slow;
}