Remove Nth Node From End of List

最新推荐文章于 2026-01-01 19:48:17 发布

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点，并返回修改后的链表头。该算法使用双指针技巧实现单次遍历，适用于各种链表操作场景。

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.

/**
 * 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* fast=head;
        ListNode* slow=head;
        ListNode* prev=NULL;
        for(int i=1;i<=n;i++){
            if(fast==NULL) return NULL;
            fast=fast->next;
        }
        //if(fast==NULL) return head;
        while(fast){
            fast=fast->next;
            prev=slow;
            slow=slow->next;
        }
        if(prev!=NULL && slow!=head)
        prev->next=slow->next;
        else head=head->next;
        return head;
    }
};