给定范围内第 K 个最大的奇数

给定两个变量L和R ，表示从L到R的整数范围，以及一个数字K ，任务是找到第 K个最大的奇数。如果K > L到R范围内的奇数个数，则返回 0。

例子：

Input: L = -10, R = 10, K = 8
Output: -5
Explanation:  The odd Numbers in the range are -9, -7, -5, -3, -1, 1, 3, 5, 7, 9 and the 8th Largest odd number is -5

Input: L = -3, R = 3, K = 1
Output: 3

方法：给定的问题可以用数学来解决。这个想法是检查R是奇数还是偶数，并相应地计算第 K 个最大的奇数。以下步骤可用于解决问题：

• 如果K<=0则返回 0
• 初始化count计算范围内奇数个数
• 如果R是奇数：
  • 计数= ceil((浮点数)(R-L+1)/2)
  • 如果K > count返回 0
• 否则返回 (R – 2*K + 2)
• 如果R是偶数
  • 计数= 楼层((R-L+1)/2)
  • 如果K > count返回 0
  • 否则返回 (R – 2*K + 1)

下面是上述方法的实现：

C++

// C++ implementation for the above approach
 
#include 
#include 
using namespace std;
 
// Function to return Kth largest
// odd number if it exists
int kthOdd(pair range, int K)
{
 
    // Base Case
    if (K <= 0)
        return 0;
 
    int L = range.first;
    int R = range.second;
 
    if (R & 1) {
 
        // Calculate count of odd
        // numbers within the range
        int Count = ceil((float)(R - L + 1) / 2);
 
        // if k > range then kth largest
        // odd number is not in the range
        if (K > Count)
            return 0;
 
        else
            return (R - 2 * K + 2);
    }
    else {
 
        // Calculate count of odd
        // numbers within the range
        int Count = (R - L + 1) / 2;
 
        // If k > range then kth largest
        // odd number is not in this range
        if (K > Count)
            return 0;
 
        else
            return (R - 2 * K + 1);
    }
}
 
// Driver Code
int main()
{
    // Initialize the range
    pair p = { -10, 10 };
 
    // Initialize k
    int k = 8;
 
    // print the kth odd number
    cout << kthOdd(p, k);
 
    return 0;
}

Java

// Java implementation for the above approach
class GFG {
 
    // Function to return Kth largest
    // odd number if it exists
    public static int kthOdd(int[] range, int K) {
 
        // Base Case
        if (K <= 0)
            return 0;
 
        int L = range[0];
        int R = range[1];
 
        if ((R & 1) > 0) {
 
            // Calculate count of odd
            // numbers within the range
            int Count = (int) Math.ceil((R - L + 1) / 2);
 
            // if k > range then kth largest
            // odd number is not in the range
            if (K > Count)
                return 0;
 
            else
                return (R - 2 * K + 2);
        } else {
 
            // Calculate count of odd
            // numbers within the range
            int Count = (R - L + 1) / 2;
 
            // If k > range then kth largest
            // odd number is not in this range
            if (K > Count)
                return 0;
 
            else
                return (R - 2 * K + 1);
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
       
        // Initialize the range
        int[] p = { -10, 10 };
 
        // Initialize k
        int k = 8;
 
        // print the kth odd number
        System.out.println(kthOdd(p, k));
    }
}
 
// This code is contributed by gfgking.

Python3

# python implementation for the above approach
import math
 
# Function to return Kth largest
# odd number if it exists
def kthOdd(range, K):
 
    # Base Case
    if (K <= 0):
        return 0
 
    L = range[0]
    R = range[1]
 
    if (R & 1):
 
        # Calculate count of odd
        # numbers within the range
        Count = math.ceil((R - L + 1) / 2)
 
        # if k > range then kth largest
        # odd number is not in the range
        if (K > Count):
            return 0
 
        else:
            return (R - 2 * K + 2)
 
    else:
 
        # Calculate count of odd
        # numbers within the range
        Count = (R - L + 1) // 2
 
        # If k > range then kth largest
        # odd number is not in this range
        if (K > Count):
            return 0
 
        else:
            return (R - 2 * K + 1)
 
# Driver Code
if __name__ == "__main__":
 
    # Initialize the range
    p = [-10, 10]
 
    # Initialize k
    k = 8
 
    # print the kth odd number
    print(kthOdd(p, k))
 
# This code is contributed by rakeshsahni

C#

// C# code for the above approach
using System;
 
public class GFG
{
 
    // Function to return Kth largest
    // odd number if it exists
    public static int kthOdd(int[] range, int K) {
  
        // Base Case
        if (K <= 0)
            return 0;
  
        int L = range[0];
        int R = range[1];
  
        if ((R & 1) > 0) {
  
            // Calculate count of odd
            // numbers within the range
            int Count = ((R - L + 1) / 2);
  
            // if k > range then kth largest
            // odd number is not in the range
            if (K > Count)
                return 0;
  
            else
                return (R - 2 * K + 2);
        } else {
  
            // Calculate count of odd
            // numbers within the range
            int Count = (R - L + 1) / 2;
  
            // If k > range then kth largest
            // odd number is not in this range
            if (K > Count)
                return 0;
  
            else
                return (R - 2 * K + 1);
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
       
        // Initialize the range
        int[] p = { -10, 10 };
  
        // Initialize k
        int k = 8;
  
        // print the kth odd number
        Console.WriteLine(kthOdd(p, k));
    }
}
 
// This code is contributed by sanjoy_62.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to return Kth largest
        // odd number if it exists
        function kthOdd(range, K) {
 
            // Base Case
            if (K <= 0)
                return 0;
 
            let L = range.first;
            let R = range.second;
 
            if (R & 1) {
 
                // Calculate count of odd
                // numbers within the range
                let Count = Math.ceil((R - L + 1) / 2);
 
                // if k > range then kth largest
                // odd number is not in the range
                if (K > Count)
                    return 0;
 
                else
                    return (R - 2 * K + 2);
            }
            else {
 
                // Calculate count of odd
                // numbers within the range
                let Count = (R - L + 1) / 2;
 
                // If k > range then kth largest
                // odd number is not in this range
                if (K > Count)
                    return 0;
 
                else
                    return (R - 2 * K + 1);
            }
        }
 
        // Driver Code
 
        // Initialize the range
        let p = { first: -10, second: 10 };
 
        // Initialize k
        let k = 8;
 
        // print the kth odd number
        document.write(kthOdd(p, k));
 
    // This code is contributed by Potta Lokesh
    

输出

-5

时间复杂度：O(1)
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