问题是删除链表的倒数第 n 个节点，并返回链表的头节点。你可以使用两个指针来实现这个目标，一个快指针和一个慢指针。首先，快指针先移动 n 步，然后两个指针同时移动，直到快指针到达链表的末尾。这时，慢指针就指向了要删除节点的前一个节点，然后你可以修改指针来完成删除操作。

以下是相应的C++代码

#include <iostream>using namespace std;struct ListNode {int val;ListNode *next;ListNode(int x) : val(x), next(NULL) {}
};class Solution {
public:ListNode* removeNthFromEnd(ListNode* head, int n) {ListNode *dummy = new ListNode(0);dummy->next = head;ListNode *fast = dummy;ListNode *slow = dummy;// Advance the fast pointer by n + 1 stepsfor (int i = 0; i <= n; i++) {fast = fast->next;}// Move both pointers until the fast pointer reaches the endwhile (fast != nullptr) {fast = fast->next;slow = slow->next;}// Remove the nth node from the endslow->next = slow->next->next;return dummy->next;  // Return the new head}
};int main() {// Example usage:ListNode *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);Solution solution;ListNode *newHead = solution.removeNthFromEnd(head, 2);// Print the modified linked listwhile (newHead != nullptr) {cout << newHead->val << " ";newHead = newHead->next;}return 0;
}