19.删除链表的倒数第-n-个结点.cpp

/*
 * @lc app=leetcode.cn id=19 lang=cpp
 *
 * [19] 删除链表的倒数第 N 个结点
 *
 * https://leetcode.cn/problems/remove-nth-node-from-end-of-list/description/
 *
 * algorithms
 * Medium (48.70%)
 * Likes:    2908
 * Dislikes: 0
 * Total Accepted:    1.5M
 * Total Submissions: 3M
 * Testcase Example:  '[1,2,3,4,5]\n2'
 *
 * 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。
 *
 *
 *
 * 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？
 *
 */
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) {}
};
// @lc code=start
/**
 * 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) {
    ListNode* dummy = new ListNode(0, head);
    ListNode* first = head;
    ListNode* second = dummy;
    for (int i = 0; i < n; i++) {
      first = first->next;
    }
    while (first != nullptr) {
      first = first->next;
      second = second->next;
    }
    second->next = second->next->next;
    ListNode* ans = dummy->next;
    delete dummy;
    return ans;
  }
};
// @lc code=end