hackerrank>Dashboard>C++>STL>Deque-STL

Double ended queue or Deque(part of C++ STL) are sequence containers with dynamic sizes that can be expanded or contracted on both ends (either its front or its back). The member functions of deque that are mainly used are:

Deque Template:

std::deque<value_type>

Declaration:

deque<int> mydeque; //Creates a double ended queue of deque of int type

Size

int length = mydeque.size(); //Gives the size of the deque

Push

mydeque.push_back(1); //Pushes element at the end
mydeque.push_front(2); //Pushes element at the beginning

Pop

mydeque.pop_back(); //Pops element from the end
mydeque.pop_front(); //Pops element from the beginning

Empty

mydeque.empty() //Returns a boolean value which tells whether the deque is empty or not

To know more about deque,
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Given a set of arrays of size
and an integer
, you have to find the maximum integer for each and every contiguous subarray of size

for each of the given arrays.

Input Format

First line of input will contain the number of test cases T. For each test case, you will be given the size of array
N and the size of subarray to be used K. This will be followed by the elements of the array
Ai.

Constraints

, where

is the
element in the array
.

Output Format

For each of the contiguous subarrays of size
of each array, you have to print the maximum integer.

Sample Input

2
5 2
3 4 6 3 4
7 4
3 4 5 8 1 4 10

Sample Output

4 6 6 4
8 8 8 10

Explanation

For the first case, the contiguous subarrays of size 2 are {3,4},{4,6},{6,3} and {3,4}. The 4 maximum elements of subarray of size 2 are: 4 6 6 4.

For the second case,the contiguous subarrays of size 4 are {3,4,5,8},{4,5,8,1},{5,8,1,4} and {8,1,4,10}. The 4 maximum element of subarray of size 4 are: 8 8 8 10.

输出每一个子集的最大值。刚开始练习STL。

#include <iostream>
#include <deque>
using namespace std;
int main()
{
int N;
cin >> N;
while(N--)
{
int n,k,com;
cin >> n >> k;
deque < int > mydeq;
deque < int > :: iterator it;
pair < int , deque <int> :: iterator > place;
place.first = 0;
com = n - k ;
while ( n-- )
{
int t;
cin >> t;
mydeq.push_back (t);
}
int t = com+1;
while ( t-- )
{
if( place.first !=0 )
place.second--;
if(place.first == 0)
{
for( it = mydeq.begin() ; it != mydeq.begin()+k ; it++ )
if( *it > place.first )
{
place.first = *it;
place.second = it;
}
}
else
{
it = mydeq.begin()+k-1;
if( *it >= place.first )
{
place.first = *it;
place.second = it;
}
}
if(t == com)
cout << place.first;
else
cout << ' ' << place.first;
if ( place.second -mydeq.begin() <= 0 )
place.first = 0;
mydeq.erase( mydeq.begin() );
}
cout<<endl;
}
return 0;
}