当前位置： 面试刷题>> 链表的中点 (经典算法题500道)
### 题目描述补充
给定一个单链表的头节点 `head`，请编写一个函数来找到该链表的中点。链表中的节点定义如下（以 Python 为例，但概念适用于所有语言）：
```python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
```
函数签名应该类似于：
```python
def findMiddle(head: ListNode) -> ListNode
```
该函数需要返回链表的中点节点。如果链表长度为偶数，则返回中间两个节点的第一个。
### 示例
假设链表为 `1 -> 2 -> 3 -> 4 -> 5`，则函数应该返回节点 `3`。
如果链表为 `1 -> 2 -> 3 -> 4`，则函数应该返回节点 `2` 或 `3`（根据实现，通常返回第一个中点）。
### PHP 示例代码
```php
val = $val;
$this->next = $next;
}
}
function findMiddle($head) {
$slow = $head;
$fast = $head;
while ($fast != null && $fast->next != null) {
$slow = $slow->next;
$fast = $fast->next->next;
}
return $slow;
}
// 示例用法
$head = new ListNode(1);
$head->next = new ListNode(2);
$head->next->next = new ListNode(3);
$head->next->next->next = new ListNode(4);
$head->next->next->next->next = new ListNode(5);
$middle = findMiddle($head);
echo $middle->val; // 输出 3
?>
```
### Python 示例代码
```python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def findMiddle(head):
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
# 示例用法
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
head.next.next.next.next = ListNode(5)
middle = findMiddle(head)
print(middle.val) # 输出 3
```
### JavaScript 示例代码
```javascript
function ListNode(val, next) {
this.val = (val===undefined ? 0 : val)
this.next = (next===undefined ? null : next)
}
function findMiddle(head) {
let slow = head;
let fast = head;
while (fast !== null && fast.next !== null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
// 示例用法
let head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
head.next.next.next = new ListNode(4);
head.next.next.next.next = new ListNode(5);
let middle = findMiddle(head);
console.log(middle.val); // 输出 3
```
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