从两个范围中选择点，以使两个范围中都没有点

给定两个段[L1，R1]和[L2，R2] ，任务是从两个范围中选择两个元素x和y (一个来自范围1，另一个来自范围2)，使得没有元素同时属于两个范围，即x属于第一范围， y属于第二范围。如果不存在这样的元素，则打印-1 。

例子：

Input: L1 = 1, R1 = 6, L2 = 3, R2 = 11
Output: 1 11
1 lies only in range [1, 6] and 11 lies only in [3, 11]

Input: L1 = 5, R1 = 10, L2 = 1, R2 = 7
Output: 1 10

方法：

• 如果L1！= L2和R1！= R2，则这些点将是min(L1，L2)和max(R1，R2) 。
• 否则只能从一个范围中选择一个点，因为该范围之一完全在另一个范围内，因此我们为该点打印-1 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to find the required points
void findPoints(int l1, int r1, int l2, int r2)
{
  
    int x = (l1 != l2) ? min(l1, l2) : -1;
    int y = (r1 != r2) ? max(r1, r2) : -1;
    cout << x << " " << y;
}
  
// Driver code
int main()
{
    int l1 = 5, r1 = 10, l2 = 1, r2 = 7;
    findPoints(l1, r1, l2, r2);
}

Java

// Java implementation of the approach
class GFG
{
      
// Function to find the required points
static void findPoints(int l1, int r1, 
                       int l2, int r2)
{
  
    int x = (l1 != l2) ? Math.min(l1, l2) : -1;
    int y = (r1 != r2) ? Math.max(r1, r2) : -1;
    System.out.println(x + " " + y);
}
  
// Driver code
public static void main(String[] args)
{
    int l1 = 5, r1 = 10, l2 = 1, r2 = 7;
    findPoints(l1, r1, l2, r2);
}
}
  
// This code is contributed by Code_Mech

Python 3

# Python3 implementation of the approach
  
# Function to find the required points
def findPoints(l1, r1, l2, r2):
  
    x = min(l1, l2) if(l1 != l2) else -1
    y = max(r1, r2) if(r1 != r2) else -1
    print(x, y)
  
# Driver code
if __name__ == "__main__":
      
    l1 = 5
    r1 = 10
    l2 = 1
    r2 = 7
    findPoints(l1, r1, l2, r2)
  
# This code is contributed by ita_c

C#

// C# implementation of the approach
using System;
  
class GFG
{
    // Function to find the required points
    static void findPoints(int l1, int r1,
                            int l2, int r2)
    {
        int x = (l1 != l2) ? Math.Min(l1, l2) : -1;
        int y = (r1 != r2) ? Math.Max(r1, r2) : -1;
        Console.WriteLine(x + " " + y);
    }
      
    // Driver code
    public static void Main()
    {
        int l1 = 5, r1 = 10, l2 = 1, r2 = 7;
        findPoints(l1, r1, l2, r2);
    }
}
  
// This code is contributed by Ryuga

输出：

1 10