题目

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

19. 删除链表的倒数第 N 个结点 - 力扣（LeetCode）

示例1

输入：head = [1,2,3,4,5], n = 2 输出：[1,2,3,5]

示例2

输入：head = [1], n = 1 输出：[]

示例3

输入：head = [1,2], n = 1 输出：[1]

提示：

链表中结点的数目为 sz 1 <= sz <= 30 0 <= Node.val <= 100 1 <= n <= sz

进阶：你能尝试使用一趟扫描实现吗？

Java解法

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if ((head == null) || n <= 0) return head;
        ListNode first = head, second = head;
        for (int i = 1; i <= n; ++i) {
            if (second == null) return head;
            second = second.next;
        }

        if (second == null) {
            first = head;
            head = head.next;
            // 切断联系;
            first.next = null;
            first = null;
            return head;
        }

        while (second.next != null) {
            first = first.next;
            second = second.next;
        }

        second = first.next;
        first.next = second.next;
        // 切断联系;
        second.next = null;
        second = null;
        return head;
    }
}

C++解法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        // 基本条件检测
        if (!head || n <= 0) return head;
        // 推进;
        ListNode *pFirst = head, *pSecond = head;
        for (int i = 1; i <= n; ++i) {
            if (!pSecond) return head;
            pSecond = pSecond->next;
        }

        // 检测;
        if (pSecond == nullptr) {
            pFirst = head, head = head->next;
            delete pFirst;
            return head;
        }

        // 继续推进
        while (pSecond->next) {
            pFirst = pFirst->next;
            pSecond = pSecond->next;
        }

        pSecond = pFirst->next;
        pFirst->next = pSecond->next;
        delete pSecond;
        return head;
    }
};

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