LintCode 174: Remove Nth Node From End of List_lintcode 174. remove nth node from end of list-优快云博客

本文介绍了一种高效算法，用于在不获取链表长度的情况下删除链表倒数第N个节点。通过使用双指针法，文章详细解释了如何定位待删除节点并进行删除操作，同时提供了三种不同的实现方式。

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Remove Nth Node From End of List
中文English
Given a linked list, remove the nth node from the end of list and return its head.

Example
Example 1:
Input: list = 1->2->3->4->5->null， n = 2
Output: 1->2->3->5->null

Example 2:
Input: list = 5->4->3->2->1->null, n = 2
Output: 5->4->3->1->null

Challenge
Can you do it without getting the length of the linked list?

Notice
The minimum number of nodes in list is n.

解法1：双指针法。
注意：
1）当删除节点是头结点情况。
2）p1应该是要删除节点的前面一个节点，不然链不起来。
代码如下:

/**
 * Definition of singly-linked-list:
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *        this->val = val;
 *        this->next = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @param n: An integer
     * @return: The head of linked list.
     */
    ListNode * removeNthFromEnd(ListNode * head, int n) {
        if (!head) return NULL;
        if (n == 0) return NULL;
        
        ListNode *p1 = head, *p2 = head;
        for (int i = 0; i < n + 1; ++i) {
            p2 = p2->next;
            if (!p2) {
                ListNode *newHead = p1->next;
                p1->next = NULL;
                return newHead;
            }
        }

        while(p2) {
            p2 = p2->next; 
            p1 = p1->next;
        }

        ListNode *deleteP = p1->next;
        p1->next = p1->next->next;
        deleteP->next = NULL;
        
        return head;
    }
};

二刷：

/**
 * 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);
        dummy->next = head;
        ListNode *fast, *slow;
        fast = head; slow = dummy;
        for (int i = 0; i < n; i++) {
            if (fast == nullptr) return nullptr;
            fast = fast->next;
        }
        while (fast != nullptr) {
            fast = fast->next;
            slow = slow->next; 
        }
        ListNode *temp = slow->next;
        slow->next = slow->next->next;
        delete temp;
        temp = 0;
        
        return dummy->next;
    }
};

三刷：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (!head || n == 0) return NULL;
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        ListNode *fast = head, *slow = head, *prev = dummy;
        
        while (fast) {
            fast = fast->next;
            n--;
            if (fast && n <= 0) {
                prev = slow;
                slow = slow->next;    
            }            
        }
        prev->next = slow->next;
        return dummy->next;
    }
};