LeetCode - 19. Remove Nth Node From End of List

19. Remove Nth Node From End of List

Problem's Link

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[b]Mean:[/b]

给定一个链表，删除倒数第n个结点.

[b]analyse:[/b]

简单粗暴 :)

[b]Time complexity: O(N)[/b]

[b]view code[/b]

/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-02-17-13.51
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);

struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

class Solution
{
public:
ListNode* removeNthFromEnd(ListNode* head, int n)
{
vector<int> list;
while(head)
{
list.push_back(head->val);
head=head->next;
}
int si=list.size();
if(si<=1) return head;
// calculate the idx
int idx=si-n;
ListNode* ret;
ListNode* tmp=ret;
bool isFirst=true;
for(int i=0; i<si; ++i)
{
if(i==idx)
continue;
if(isFirst)
{
tmp=new ListNode(list[i]);
ret=tmp;
isFirst=false;
}
else
{
tmp->next=new ListNode(list[i]);
tmp=tmp->next;
}
}
return ret;
}
};

int main()
{
Solution solution;
int n,num,tmp;
while(cin>>n>>num)
{
ListNode* head=NULL,*root=NULL;
for(int i=0; i<n; ++i)
{
cin>>tmp;
if(i==0)
{
head=new ListNode(tmp);
root=head;
}
else
{
head->next=new ListNode(tmp);
head=head->next;
}
}

ListNode* ans=solution.removeNthFromEnd(root,num);
while(ans)
{
cout<<ans->val<<" ";
ans=ans->next;
}
cout<<endl;
cout<<"End."<<endl;
}
return 0;
}
/*

*/