【华为外包面试】删除链表的倒数第 n 个结点，并且返回链表的头结点

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。
例如:
输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]
提示：
链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

/**
 * 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) {
    
    

    }
};

代码：

#include <iostream>

struct LinkNode {
    
    
    int data;
    LinkNode* pNext;
};

LinkNode* GetLastNNode(LinkNode* head,int n) {
    
    
    if (head==nullptr||n<=0)
    {
    
    
        return nullptr;
    }
    int numOfList = 0;
    LinkNode* p = head;
    while (p!=nullptr)
    {
    
    
        p = p->pNext;
        numOfList++;
    }
    if (n > numOfList) {
    
    
        // 需要删除的节点不在链表里
        return head;
    }
    else if(n==numOfList)
    {
    
    
        return head->pNext;
    }
    int toDelete = numOfList - n-1;
    p = head;
    while (toDelete>0)
    {
    
    
        p = p->pNext;
    }
    //LinkNode* toDelete = p->pNext;
    p->pNext = p->pNext->pNext;

    return head;
}

int main()
{
    
    
    LinkNode* head= (LinkNode*)malloc(sizeof(LinkNode));
    head->data = 0;
    head->pNext = nullptr;
    LinkNode* iter = head;
    int length = 6;
    for (int i = 1; i < length; i++)
    {
    
    
        LinkNode *p = (LinkNode*)malloc(sizeof(LinkNode));
        p->data = i;
        p->pNext = nullptr;
        iter->pNext = p;
        iter = iter->pNext;
    }
    LinkNode* result = GetLastNNode(head->pNext, 4);
    while (result!=nullptr)
    {
    
    
        std::cout << result->data << std::endl;
        result = result->pNext;
    }
    return 0;
}