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.
题目
返回链表中间结点，若长度为偶数，则返回靠后的中间结点
思路
经典题目，快慢指针法
定义两个指针fast和slow，从head开始fast每次走两步，slow每次走一步，当fast走到头时，slow刚好走到中间，返回slow
代码
public ListNode middleNode(ListNode head) {
if (head==null||head.next==null) {//当链表为空或者只有一个元素时
return head;
}
if (head.next.next==null) {//当链表有两个元素时
return head.next;
}
ListNode fast=head;
ListNode slow=head;
while (fast!=null&&fast.next!=null) {//一般情况
fast=fast.next.next;
slow=slow.next;
}
return slow;
}
