我想输出 1 4 2 3 5 奇数升序和偶数降序

Question

我想按升序输出奇数，按降序输出偶数我的代码在这里。

当我给输入n=5数组5 2 4 3 1，我想输出1 4 2 3 5， 但我得到了输出4 2 1 3 5。我不想要数组的位置。

#include <iostream>
using namespace std;

int main()
{
    int n;
    cin >> n;
    int arr[n];

    for (int i = 0; i < n; ++i)
    {
        cin >> arr[i];
    }

// odd number ascending order
    for (int i = 0; i <= n; ++i)
    {
        for (int j = i + 1; j < n; ++j)
        {
            if (arr[i] < arr[j])
            {
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
    }
    for (int i = 0; i < n; ++i)
    {
        if (arr[i] % 2 == 0)
        {
            cout << arr[i]<< " ";
        }
    }

// numbers in descending order
    for (int i = 0; i <= n - 1; i++)
    {
        for (int j = i + 1; j <= n; j++)
        {

            if (arr[j] < arr[i])
            {
                int temp = arr[j];
                arr[j] = arr[i];
                arr[i] = temp;
            }
        }
        if (arr[i] % 2 == 1)
        {
            cout << arr[i]<<" ";
        }
    }

    return 0;
}

Answer 1

我们，初学者，应该互相帮助。:)

对于你我这样的初学者来说，这项任务并不容易。

我可以建议一种基于冒泡排序方法的方法。

你来了。

#include <iostream>
#include <utility>

template <typename UnaryPredicate>
void conditional_bubble_sort( int a[], size_t n, 
                              UnaryPredicate &&unary_predicate,
                              bool order = false )
{
    while ( n && !unary_predicate( *a ) )
    {
        ++a;
        --n;
    }
    
    for ( size_t i = 0, last = 0; !( n < 2 ); n = last )
    {
        last = 0;
        
        for ( size_t current = 0, next = 0; next < n; current = next )
        {
            while ( ++next < n && !unary_predicate( a[next] ) );
            
            if ( next != n )
            {
                if ( !order /* ascending */ ? a[next] < a[current] : a[current] < a[next] )
                {
                    std::swap( a[current], a[next] );
                    last = next;
                }
            }
        }
    }
}

int main() 
{
    int a[] = { 5, 2, 4, 3, 1 };
    const size_t N = sizeof( a ) / sizeof( *a );
    
    for ( const auto &item : a )
    {
        std::cout << item << ' ';
    }
    std::cout << '\n';
    
    conditional_bubble_sort( a, N, []( const auto &item ) { return item % 2 != 0;} );
    
    for ( const auto &item : a )
    {
        std::cout << item << ' ';
    }
    std::cout << '\n';
    
    conditional_bubble_sort( a, N, []( const auto &item ) { return item % 2 == 0;}, true );
    
    for ( const auto &item : a )
    {
        std::cout << item << ' ';
    }
    std::cout << '\n';
    
    return 0;
}

程序输出是

5 2 4 3 1 
1 2 4 3 5 
1 4 2 3 5

Answer 2

#include <iostream>
using namespace std;

int main()
{
    int n = 5;
    int left = 1, right = n - (n % 2);
    for(int i = 1; i <= n; i++)
    {
        if(i % 2 == 0)
        {
            cout << right << ' ';
            right-= 2;
        }
        else if(i % 2 == 1)
        {
            cout << left << ' ';
            left += 2;
        }
    }
    return 0;
}