174. Remove Nth Node From End of List 删除倒数第N个节点

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) {
        if (head == nullptr || head->next == nullptr)
            return nullptr;
        
        ListNode *pre = head;
        ListNode *cur = nullptr;
        
        for (int i = 0; i < n-1; i++)
            if (pre->next != nullptr)
            {
                pre = pre->next;
            }
            else
            {
                return nullptr;
            }
            
        pre = pre->next;
        if (pre == nullptr)
        return head->next;
        
        cur = head;
        while(pre->next != nullptr)
        {
            pre = pre->next;
            cur = cur->next;
        }
        
        cur->next = cur->next->next;
        return head;
    }
};

 

/**
 * 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 == nullptr || head->next == nullptr)
            return nullptr;
        
        ListNode *pre = head;
        ListNode *cur = nullptr;
        
        for (int i = 0; i < n; i++)
            if (pre != nullptr)
            {
                pre = pre->next;
            }
            else
            {
                return nullptr;
            }
            
        
        if (pre == nullptr)
        return head->next;
        
        cur = head;
        while(pre->next != nullptr)
        {
            pre = pre->next;
            cur = cur->next;
        }
        
        cur->next = cur->next->next;
        return head;
    }
};

 

// 面试题22：链表中倒数第k个结点
// 题目：输入一个链表，输出该链表中倒数第k个结点。为了符合大多数人的习惯，
// 本题从1开始计数，即链表的尾结点是倒数第1个结点。例如一个链表有6个结点，
// 从头结点开始它们的值依次是1、2、3、4、5、6。这个链表的倒数第3个结点是
// 值为4的结点。

#include <cstdio>
#include "..\Utilities\List.h"

ListNode* FindKthToTail(ListNode* pListHead, unsigned int k)
{
    if(pListHead == nullptr || k == 0)
        return nullptr;

    ListNode *pAhead = pListHead;
    ListNode *pBehind = nullptr;

    for(unsigned int i = 0; i < k - 1; ++ i)
    {
        if(pAhead->m_pNext != nullptr)
            pAhead = pAhead->m_pNext;
        else
        {
            return nullptr;
        }
    }

    pBehind = pListHead;
    while(pAhead->m_pNext != nullptr)
    {
        pAhead = pAhead->m_pNext;
        pBehind = pBehind->m_pNext;
    }

    return pBehind;
}