[leetcode]Remove Nth Node From End of List

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本文介绍了一种高效算法，用于从链表中移除倒数第N个节点。通过使用双指针技巧，可以在一次遍历内完成操作。提供了三种实现方法，包括双指针保持N个单位的距离、双指针相隔N-1逐个后移以及添加虚拟头节点的方法。

题目来源：https://leetcode.com/problems/remove-nth-node-from-end-of-list/

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.

方法1：双指针，距离n。

/**
 * 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) {
        ListNode *pre, *post;
        pre = post = head;
        while(n--) post = post->next;
        if(!post) {
            ListNode* t = head;
            head = head->next;
            delete t;
            return head;
        }
        while(post->next) {
            pre = pre->next;
            post = post->next;
        }
        ListNode* t = pre->next;
        pre->next = t->next;
        delete t;
        return head;
    }
};

方法2：两个指针相隔n-1，每次两个指针向后一步，当后面一个指针没有后继了，前面一个指针就是要删除的节点。

/**
 * 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;
            
        ListNode *pPre = NULL;
        ListNode *p = head;
        ListNode *q = head;
        for(int i = 0; i < n - 1; i++)
            q = q->next;
            
        while(q->next)
        {
            pPre = p;
            p = p->next;
            q = q->next;
        }
        
        if (pPre == NULL)
        {
            head = p->next;
            delete p;
        }
        else
        {
            pPre->next = p->next;
            delete p;
        }
        
        return head;
    }
};

方法3：增加一个节点，最头节点。

#include <iostream>
using namespace std;
struct LinkNode
{
	int data;
	LinkNode *next;
	LinkNode(int x):data(x),next(NULL){}
};

LinkNode * removeNthNodeFromEnd(LinkNode *head,int n)
{
	if(head == NULL || n == 0)//防御性编程，增加健壮性
		return head;
	LinkNode *dummy = new LinkNode(-1);
	dummy->next = head;
	LinkNode *p,*q,*tmp;
	p = dummy;
	q = dummy;
	
	//p先走n步
	for(int i=0;i<n;i++)
	{
		if(p->next != NULL)
			p = p->next;
		else
			return head;//如果n超过链表的长度，直接返回head
	}

	//然后p,q一起走
	while(p->next)
	{
		p = p->next;
		q = q->next;
	}

	//此时q指向的是待删除节点的前驱
	tmp = q->next->next;
	delete q->next;
	q->next = tmp;

	head = dummy->next;
	delete dummy;
	return head;
}