如何在c++中仅使用if和else if对3个变量进行升序排序

Question

正如标题所说，我需要对3个数字进行升序排序，而不使用无效交换等。我可以使用else，如果else.The 3数字不需要不同，例如，我们可以使用: a -5，b -5，c-10，输出应该是-5-5 10。我这样做是为了一个在线学习程序，但我的代码得到了100分中的93分，因此我不能跳到下一课。这是我的代码：

#include <iostream>
using namespace std;

int main() {
    int a, b, c;
    cin >> a >> b >> c;

    if ( a < b && b < c ) {
        cout << a << " " << b << " " << c;
    } else if ( a > b && b > c ) {
        cout << c << " " << b << " " << a;
    }

    if ( a < c && b < a ) {
    cout << b << " " << a << " " << c;
    }

    if ( a < c && c < b ) {
        cout << a << " " << c << " " << b;
    } else if ( a > c && b > a ) {
        cout << c << " " << a << " " << b;
    }

    if ( a == b && b == c ) {
        cout << a << " " << b << " " << c;
    } else if ( a == c && b == a ) {
        cout << b << " " << c << " " << a;
    }

    if ( a == b && c < a ) {
        cout << c << " " << b << " " << a;
    } else if ( a == b && c > a ) {
        cout << b << " " << a << " " << c;
    }

    if ( a == c && b < a ) {
        cout << b << " " << c << " " << a;
    } else if ( a == c && b > a ) {
        cout << a << " " << c << " " << b;
    }

    if ( b == c && a < b ) {
        cout << a << " " << b << " " << c;
    } else if ( b == c && a > b ) {
        cout << b << " " << c << " " << a;
    }

   return 0;
}

提前谢谢你。

Answer 1

使用min/max函数的值较小，但我觉得可能有更好的方法来找到的中间值。

int min = std::min(a, std::min(b, c));
int max = std::max(a, std::max(b, c));

std::cout << min << " ";

int middle = a; // By default a, in case none of the following conditions is true
if (a != min && a != max)
    middle = a;
else if (b != min && b != max)
    middle = b;
else if (c != min && c != max)
    middle = c;

std::cout << middle << " " << max;

Answer 2

这样如何：

int x[3];
// first we compute the smallest
x[0] = min(a, min(b, c));

// then the biggest
x[2] = max(a, max(b, c));

// finally we take the remaining one as the middle element. For that we use this substraction trick
x[1] = a+b+c - x[0]-[2];
// NOTE: if you are worried about overflow, you can use the following XOR hack:
// x[1] = a^b^c ^ x[0]^x[2];
// unlike with +/-, this is well defined behaviour (note that this is a problem only for `int`, `unsigned` overflow is well defined, so it would work even it overflows)

cout << x[0] << " " << x[1] << " " << x[2] << endl;

Answer 3

这可以使用下面的if-else语句序列来完成。

if ( !( b < a ) && !( c < a ) )
{
    if ( !( c < b ) )
    {
        std::cout << a << ' ' << b << ' ' << c << '\n';
    }
    else
    {
        std::cout << a << ' ' << c << ' ' << b << '\n';
    } 
}
else if ( !( a < b ) && !( c < b ) )
{
    if ( !( c < a ) )
    {
        std::cout << b << ' ' << a << ' ' << c << '\n';
    }
    else
    {
        std::cout << b << ' ' << c << ' ' << a << '\n';
    } 
}
else if ( !( a < c ) && !( b < c ) )
{
    if ( !( b < a ) )
    {
        std::cout << c << ' ' << a << ' ' << b << '\n';
    }
    else
    {
        std::cout << c << ' ' << b << ' ' << a << '\n';
    } 
}