标签：note serial color 中间 指针 and out public node
Easy
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:
1 and 100.题目大意：
给出一个链表，总中间截取链表并取后半部分。
方法一：先算出链表的长度，然后返回后半部分链表
代码如下：
class Solution { public: ListNode* middleNode(ListNode* head) { ListNode *dummy = head; int len = 0; while (dummy) { dummy = dummy->next; len++; } len = len / 2; while (len > 0) { head = head->next; len--; } return head;
} };
方法二：快慢指针。找中间值，就让快指针每次移动2步，慢指针一次移动1步，直到快指针为空（链表长度为偶数）或快指针的next为空（链表长度为奇数）为止，此时慢指针的链表就是目标结果。
代码如下：
class Solution { public: ListNode* middleNode(ListNode* head) { ListNode *slow = head, *fast = head; while (fast&&fast->next) { slow = slow->next; fast = fast->next->next; } return slow; } };
[LeetCode] 876. Middle of the Linked List
标签：note serial color 中间 指针 and out public node
原文地址：https://www.cnblogs.com/cff2121/p/11646141.html
