leetcode19.删除链表的倒数第 N 个结点

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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：

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

分析：

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快慢指针，快指针先走n，慢指针开始走，当快指针走到尾部时，慢指针走到倒数第n个

代码

  • C++

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/**
 * 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* fast = head, *slow = head;
        for (int i = 0;i < n;i ++)
            fast = fast->next;
        
        if (fast == nullptr)
            return head->next;

        while (fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next;
        }
        slow->next = slow->next->next;
        return head;
    }
};

[原题链接](19. 删除链表的倒数第 N 个结点 - 力扣（Leetcode）)

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