给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。
示例：
给定一个链表: 1->2->3->4->5, 和 n = 2.
当删除了倒数第二个节点后，链表变为 1->2->3->5.
说明：
给定的 n 保证是有效的。
进阶：
你能尝试使用一趟扫描实现吗？
题解
递归法
利用递归的特性，从递归栈底向上累加即为倒数索引。
removeNode 最终返回的值为输入链表的长度，若该返回值与 n 相等，则说明去掉的是第一个节点，此时头节点为 head.next 。
Java
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| class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
}
}
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
return removeNode(head, n) == n ? head.next : head;
}
public int removeNode(ListNode node, int n) {
if (node.next == null) return 1;
int m = removeNode(node.next, n);
if (m == n)
node.next = node.next.next;
return m + 1;
}
}
|
Python
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| class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
return head.next if self.remove_node(head, n) == n else head
def remove_node(self, head: ListNode, n: int) -> int:
if not head.next: return 1
m = self.remove_node(head.next, n)
if m == n:
head.next = head.next.next
return m + 1
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Go
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| package main
type ListNode struct {
Val int
Next *ListNode
}
func removeNthFromEnd(head *ListNode, n int) *ListNode {
if removeNode(head, n) == n {
return head.Next
} else {
return head
}
}
func removeNode(head *ListNode, n int) int {
if head.Next == nil {
return 1
}
m := removeNode(head.Next, n)
if m == n {
head.Next = head.Next.Next
}
return m + 1
}
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双指针法
使用快慢指针来查找索引，快指针先移动 n 个位置，再同时移动两指针，当快指针指向链表的尾节点时，慢指针所指即为倒数第 n 个节点。
Java
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| class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
if (head == null || head.next == null) return null;
ListNode fast = head, slow = head;
for (int i = 0; i < n; i++)
fast = fast.next;
if (fast == null)
return head.next;
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return head;
}
}
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Python
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| class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
fast, slow = head, head
for i in range(n):
fast = fast.next
if not fast: return head.next
while fast.next:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return head
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Go
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| func removeNthFromEnd(head *ListNode, n int) *ListNode {
fast, slow := head, head
for i := 0; i < n; i++ {
fast = fast.Next
}
if fast == nil {
return head.Next
}
for fast.Next != nil {
fast = fast.Next
slow = slow.Next
}
slow.Next = slow.Next.Next
return head
}
|