23. Merge k Sorted Lists
Problem's Link
----------------------------------------------------------------------------
Mean:
将k个有序链表合并为一个有序链表.
analyse:
方法一：本着不重复发明轮子的原则，使用两两合并，就可以利用前面已经做过的Merge two Sorted Lists.
方法二：抖机灵.将所有结点的val都push_back在一个vector容器中，排序后又重新构造链表.
Time complexity: O(N)
view code
方法一：
ListNode
*
mergeKLists(
vector
<
ListNode
*>
&
lists)
{
if( lists . empty ()){
return nullptr;
}
while( lists . size() > 1 ){
lists . push_back( mergeTwoLists( lists [ 0 ], lists [ 1 ]));
lists . erase( lists . begin());
lists . erase( lists . begin());
}
return lists . front();
}
ListNode * mergeTwoLists( ListNode * l1 , ListNode * l2) {
if( l1 == nullptr ){
return l2;
}
if( l2 == nullptr ){
return l1;
}
if( l1 -> val <= l2 -> val ){
l1 -> next = mergeTwoLists( l1 -> next , l2);
return l1;
}
else {
l2 -> next = mergeTwoLists( l1 , l2 -> next);
return l2;
}
}
if( lists . empty ()){
return nullptr;
}
while( lists . size() > 1 ){
lists . push_back( mergeTwoLists( lists [ 0 ], lists [ 1 ]));
lists . erase( lists . begin());
lists . erase( lists . begin());
}
return lists . front();
}
ListNode * mergeTwoLists( ListNode * l1 , ListNode * l2) {
if( l1 == nullptr ){
return l2;
}
if( l2 == nullptr ){
return l1;
}
if( l1 -> val <= l2 -> val ){
l1 -> next = mergeTwoLists( l1 -> next , l2);
return l1;
}
else {
l2 -> next = mergeTwoLists( l1 , l2 -> next);
return l2;
}
}
方法二：
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-02-18-09.02
*/
#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 * mergeKLists( vector < ListNode *>& lists)
{
vector < int > nodeVal;
for( auto ptr: lists)
{
while( ptr)
{
nodeVal . push_back( ptr -> val);
ptr = ptr -> next;
}
}
if( nodeVal . size() <= 0)
{
return nullptr;
}
sort( nodeVal . begin (), nodeVal . end());
bool isFirst = 1;
ListNode * res = nullptr , * head = nullptr;
for( auto p: nodeVal)
{
if( isFirst)
{
isFirst = 0;
head = new ListNode(p);
res = head;
}
else
{
head -> next = new ListNode(p);
head = head -> next;
}
}
return res;
}
};
ListNode * getList( vector < int >& arr)
{
bool isFirst = 1;
ListNode * res = nullptr , * head = nullptr;
for( auto p: arr)
{
if( isFirst)
{
isFirst = 0;
head = new ListNode(p);
res = head;
}
else
{
head -> next = new ListNode(p);
head = head -> next;
}
}
return res;
}
int main()
{
Solution solution;
int n;
while( cin >>n)
{
vector < ListNode *> listVe;
while(n --)
{
int num;
cin >> num;
vector < int > ve;
for( int i = 0; i < num; ++ i)
{
int tmp;
cin >> tmp;
ve . push_back( tmp);
}
listVe . push_back( getList( ve));
}
ListNode * head = solution . mergeKLists( listVe);
while( head)
{
cout << head -> val << " ";
head = head -> next;
}
cout << "End." << endl;
}
return 0;
}
/*
*/
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-02-18-09.02
*/
#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 * mergeKLists( vector < ListNode *>& lists)
{
vector < int > nodeVal;
for( auto ptr: lists)
{
while( ptr)
{
nodeVal . push_back( ptr -> val);
ptr = ptr -> next;
}
}
if( nodeVal . size() <= 0)
{
return nullptr;
}
sort( nodeVal . begin (), nodeVal . end());
bool isFirst = 1;
ListNode * res = nullptr , * head = nullptr;
for( auto p: nodeVal)
{
if( isFirst)
{
isFirst = 0;
head = new ListNode(p);
res = head;
}
else
{
head -> next = new ListNode(p);
head = head -> next;
}
}
return res;
}
};
ListNode * getList( vector < int >& arr)
{
bool isFirst = 1;
ListNode * res = nullptr , * head = nullptr;
for( auto p: arr)
{
if( isFirst)
{
isFirst = 0;
head = new ListNode(p);
res = head;
}
else
{
head -> next = new ListNode(p);
head = head -> next;
}
}
return res;
}
int main()
{
Solution solution;
int n;
while( cin >>n)
{
vector < ListNode *> listVe;
while(n --)
{
int num;
cin >> num;
vector < int > ve;
for( int i = 0; i < num; ++ i)
{
int tmp;
cin >> tmp;
ve . push_back( tmp);
}
listVe . push_back( getList( ve));
}
ListNode * head = solution . mergeKLists( listVe);
while( head)
{
cout << head -> val << " ";
head = head -> next;
}
cout << "End." << endl;
}
return 0;
}
/*
*/
