25. Reverse Nodes in k-Group
Problem's Link
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Mean:
给定一个链表和一个k值,将链表按照k个结点为一组,组内翻转.
analyse:
继续抖机灵!
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-02-19-11.06
*/
#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);
// Definition for singly-linked list.
struct ListNode
{
int val;
ListNode * next;
ListNode( int x) : val( x ), next( NULL) {}
};
class Solution
{
public :
ListNode * reverseKGroup( ListNode * head , int k)
{
vector < int > ve;
ListNode * tptr = head;
while( head)
{
ve . push_back( head -> val);
head = head -> next;
}
if( ve . size() < k) return tptr;
delete( head);
int frontIndex = 0 , backIndex = k - 1;
while( frontIndex < ve . size() && backIndex < ve . size())
{
int low = frontIndex , high = backIndex;
while( low < high)
{
swap( ve [ low ], ve [ high ]);
++ low , -- high;
}
frontIndex = backIndex + 1;
backIndex += k;
}
int isFirst = 1;
ListNode * res = nullptr , *p = nullptr;
for( int i = 0; i < ve . size(); ++ i)
{
if( isFirst)
{
isFirst = 0;
p = new ListNode( ve [ i ]);
res =p;
}
else
{
p -> next = new ListNode( ve [ i ]);
p =p -> next;
}
}
return res;
}
};
int main()
{
Solution solution;
int n , k;
while( cin >>n >> k)
{
bool isFirst = 1;
ListNode * res = nullptr , * head = nullptr;
for( int i = 0; i <n; ++ i)
{
int tmp;
cin >> tmp;
if( isFirst)
{
isFirst = 0;
head = new ListNode( tmp);
res = head;
}
else
{
head -> next = new ListNode( tmp);
head = head -> next;
}
}
ListNode * ans = solution . reverseKGroup( res , k);
while( ans)
{
cout << ans -> val << " ";
ans = ans -> next;
}
cout << "End." << endl;
}
return 0;
}
/*
*/
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-02-19-11.06
*/
#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);
// Definition for singly-linked list.
struct ListNode
{
int val;
ListNode * next;
ListNode( int x) : val( x ), next( NULL) {}
};
class Solution
{
public :
ListNode * reverseKGroup( ListNode * head , int k)
{
vector < int > ve;
ListNode * tptr = head;
while( head)
{
ve . push_back( head -> val);
head = head -> next;
}
if( ve . size() < k) return tptr;
delete( head);
int frontIndex = 0 , backIndex = k - 1;
while( frontIndex < ve . size() && backIndex < ve . size())
{
int low = frontIndex , high = backIndex;
while( low < high)
{
swap( ve [ low ], ve [ high ]);
++ low , -- high;
}
frontIndex = backIndex + 1;
backIndex += k;
}
int isFirst = 1;
ListNode * res = nullptr , *p = nullptr;
for( int i = 0; i < ve . size(); ++ i)
{
if( isFirst)
{
isFirst = 0;
p = new ListNode( ve [ i ]);
res =p;
}
else
{
p -> next = new ListNode( ve [ i ]);
p =p -> next;
}
}
return res;
}
};
int main()
{
Solution solution;
int n , k;
while( cin >>n >> k)
{
bool isFirst = 1;
ListNode * res = nullptr , * head = nullptr;
for( int i = 0; i <n; ++ i)
{
int tmp;
cin >> tmp;
if( isFirst)
{
isFirst = 0;
head = new ListNode( tmp);
res = head;
}
else
{
head -> next = new ListNode( tmp);
head = head -> next;
}
}
ListNode * ans = solution . reverseKGroup( res , k);
while( ans)
{
cout << ans -> val << " ";
ans = ans -> next;
}
cout << "End." << endl;
}
return 0;
}
/*
*/