从两个范围中选择点，使得两个范围内都没有点

给定两个片段[L1, R1]和[L2, R2] ，任务是从两个范围（一个来自范围一，另一个来自范围二）中选择两个元素x和y ，这样没有元素属于这两个范围，即x属于第一个范围， y属于第二个范围。如果不存在这样的元素，则改为打印-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那么点将是min(L1, L2)和max(R1, R2) 。
• 否则只能从一个范围中选择一个点，因为其中一个范围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

Python3

# 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

Javascript

 
// Javascript implementation of the approach   
 
// Function to find the required points
function findPoints(l1 , r1 , l2 , r2)
{
    var x = (l1 != l2) ? Math.min(l1, l2) : -1;
    var y = (r1 != r2) ? Math.max(r1, r2) : -1;
    document.write(x + " " + y);
}
 
// Driver code
var l1 = 5, r1 = 10, l2 = 1, r2 = 7;
 
findPoints(l1, r1, l2, r2);
 
// This code is contributed by Rajput-Ji
 

输出：

1 10

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