当前位置： 面试刷题>> 链表倒数第n个节点 (经典算法题500道)
### 题目描述
给定一个单链表，要求找到该链表中倒数第n个节点。如果链表的长度小于n，则返回null（或在特定语言中为None/null/undefined等，表示不存在）。
### 示例
假设链表是 `1 -> 2 -> 3 -> 4 -> 5`，且 n = 2。
则倒数第n个节点是 `4`。
### PHP 代码示例
```php
val = $val;
$this->next = $next;
}
}
function findNthFromEnd($head, $n) {
$slow = $head;
$fast = $head;
// 先让快指针向前走n步
for ($i = 0; $i < $n; $i++) {
if ($fast == null) {
// 链表长度小于n
return null;
}
$fast = $fast->next;
}
// 快慢指针同时移动，直到快指针到达末尾
while ($fast != null) {
$slow = $slow->next;
$fast = $fast->next;
}
// 慢指针此时指向倒数第n个节点
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);
$n = 2;
$result = findNthFromEnd($head, $n);
if ($result) {
echo "倒数第{$n}个节点的值为: " . $result->val;
} else {
echo "链表长度小于{$n}，不存在倒数第{$n}个节点";
}
?>
```
### Python 代码示例
```python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def findNthFromEnd(head: ListNode, n: int) -> ListNode:
slow = fast = head
for _ in range(n):
if not fast:
return None
fast = fast.next
while fast:
slow = slow.next
fast = fast.next
return slow
# 示例用法
head = ListNode(1)
head.next = ListNode(2)
head.next.next = ListNode(3)
head.next.next.next = ListNode(4)
head.next.next.next.next = ListNode(5)
n = 2
result = findNthFromEnd(head, n)
if result:
print(f"倒数第{n}个节点的值为: {result.val}")
else:
print(f"链表长度小于{n}，不存在倒数第{n}个节点")
```
### JavaScript 代码示例
```javascript
function ListNode(val, next) {
this.val = (val===undefined ? 0 : val)
this.next = (next===undefined ? null : next)
}
function findNthFromEnd(head, n) {
let slow = head;
let fast = head;
// 先让快指针向前走n步
for (let i = 0; i < n; i++) {
if (!fast) return null;
fast = fast.next;
}
// 快慢指针同时移动，直到快指针到达末尾
while (fast) {
slow = slow.next;
fast = fast.next;
}
// 慢指针此时指向倒数第n个节点
return slow;
}
// 示例用法
let head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
head.next.next.next = new ListNode(4);
head.next.next.next.next = new ListNode(5);
let n = 2;
let result = findNthFromEnd(head, n);
if (result) {
console.log(`倒数第${n}个节点的值为: ${result.val}`);
} else {
console.log(`链表长度小于${n}，不存在倒数第${n}个节点`);
}
```
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