LeetCode | 19. Remove Nth Node From End of List-优快云博客

本文链接：https://blog.youkuaiyun.com/pku_Coder/article/details/70553357

本文介绍了一种高效算法，用于删除单链表中倒数第N个节点，仅通过一次遍历实现。提供了两种解决方案：一种是先计算链表长度，再定位到目标节点；另一种使用快慢指针技巧直接找到并删除目标节点。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a linked list, remove the nth node from the end of list and return its head.

For example,

 Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

其实就是删除链表倒数第N个节点.注意链表无头结点,head就是第一个节点.

//自己的代码,9ms AC
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int tot = 1, i = 1;
        ListNode* it = head;
        while(it->next)
        {
            tot++;
            it = it->next;
        }
        it = head;//根据题意,head就是第一个元素
        if(n == tot)//删除第一个元素
        {
            head = head->next;
            return head;
        }
        while(i<tot-n)
        {
            i++;
            it = it->next;
        }
        if(n == 1)//删除最后一个
        {
            it->next = NULL;
            return head;
        }
        ListNode* it2 = it->next;
        ListNode* it3 = it2->next;
        it->next = it3;
        it2->next = NULL;
        delete(it2);
        return head;
    }
};

//后面提供的代码,用快指针和慢指针,只需遍历一遍.9ms AC
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head == NULL)
        return NULL;

        ListNode new_head(-1);
        new_head.next = head;

        ListNode *slow = &new_head, *fast = &new_head;

        for (int i = 0; i < n; i++)
            fast = fast->next;

        while (fast->next) 
        {
            fast = fast->next;
            slow = slow->next;
        }

        ListNode *to_be_deleted = slow->next;   //待删指针为slow->next
        slow->next = slow->next->next;

        delete to_be_deleted;

        return new_head.next;
    }
};