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.

 

由于删除头结点时要改变头结点的指针，所以传入的是引用或者使用二级指针。提示中给的只是传入一级指针，要是删除头结点，始终不能删掉。（注意要改变一个数据结构时，要传二级指针或者引用，树中的很多操作都是这样的）

C++实现代码如下：

#include<iostream>
#include<new>
using namespace std;

//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 *p=head;
        ListNode *q=head;
        ListNode *pre=head;
        int i=0;
        while(q&&i<n)
        {
            q=q->next;
            i++;
        }
        //说明没有n个元素
        if(i<n)
           return head;
        while(q)
        {
            pre=p;
            p=p->next;
            q=q->next;
        }
        if(p==head)
        {
            head=head->next;
            delete p;
        }
        else
        {
            pre->next=p->next;
            delete p;
        }
        return head;
    }
    void createList(ListNode *&head)
    {
        ListNode *p=NULL;
        int i=0;
        int arr[10]= {10,9,8,7,6,5,4,3,2,1};
        for(i=0; i<5; i++)
        {
            if(head==NULL)
            {
                head=new ListNode(arr[i]);
                if(head==NULL)
                    return;
            }
            else
            {
                p=new ListNode(arr[i]);
                p->next=head;
                head=p;
            }
        }
    }
};

int main()
{
    Solution s;
    ListNode *L=NULL;
    s.createList(L);
    ListNode *head=L;
    while(head)
    {
        cout<<head->val<<" ";
        head=head->next;
    }
    cout<<endl;
    s.removeNthFromEnd(L,5);
    while(L)
    {
        cout<<L->val<<" ";
        L=L->next;
    }
}

运行结果：

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