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82 lines
1.8 KiB
Markdown
82 lines
1.8 KiB
Markdown
# [876. Middle of the Linked List](https://leetcode.com/problems/middle-of-the-linked-list/)
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## 题目
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Given a non-empty, singly linked list with head node head, return a middle node of linked list.
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If there are two middle nodes, return the second middle node.
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**Example 1**:
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```
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Input: [1,2,3,4,5]
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Output: Node 3 from this list (Serialization: [3,4,5])
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The returned node has value 3. (The judge's serialization of this node is [3,4,5]).
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Note that we returned a ListNode object ans, such that:
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ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
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```
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**Example 2**:
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```
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Input: [1,2,3,4,5,6]
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Output: Node 4 from this list (Serialization: [4,5,6])
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Since the list has two middle nodes with values 3 and 4, we return the second one.
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```
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**Note**:
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- The number of nodes in the given list will be between 1 and 100.
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## 题目大意
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输出链表中间结点。这题在前面题目中反复出现了很多次了。
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如果链表长度是奇数，输出中间结点是中间结点。如果链表长度是双数，输出中间结点是中位数后面的那个结点。
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## 解题思路
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这道题有一个很简单的做法，用 2 个指针只遍历一次就可以找到中间节点。一个指针每次移动 2 步，另外一个指针每次移动 1 步，当快的指针走到终点的时候，慢的指针就是中间节点。
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## 代码
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```go
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package leetcode
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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func middleNode(head *ListNode) *ListNode {
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if head == nil || head.Next == nil {
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return head
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}
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p1 := head
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p2 := head
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for p2.Next != nil && p2.Next.Next != nil {
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p1 = p1.Next
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p2 = p2.Next.Next
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}
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length := 0
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cur := head
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for cur != nil {
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length++
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cur = cur.Next
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}
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if length%2 == 0 {
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return p1.Next
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}
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return p1
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}
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``` |