LeetCode-Remove Nth Node From End of List

最新推荐文章于 2025-09-11 11:17:46 发布

转载最新推荐文章于 2025-09-11 11:17:46 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/superzrx/p/3269280.html

文章标签：

#c/c++

本文介绍了一种高效的方法来删除链表中倒数第N个节点，并提供了详细的C++实现代码。该方法仅需遍历一次链表，通过记录节点地址的方式找到待删除节点，并进行相应的指针调整。

Given a linked list, remove the n^th 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.

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(head==NULL)return NULL;
        vector<ListNode*> record;
        ListNode* ptr=head;
        while(ptr!=NULL){
            record.push_back(ptr);
            ptr=ptr->next;
        }
        int total=record.size();
        if(record.size()==n)return head->next;
        else if(n==1){
            record[total-2]->next=NULL;
        }
        else{
           
            record[total-n-1]->next=record[total-n+1];
        }
        return head;
    }
};

转载于:https://www.cnblogs.com/superzrx/p/3269280.html