给定范围内第 k 个最大奇数
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 Kth 最大奇数。如果 K > 的奇数个数在 L 到 R 的范围内，则返回 0。
示例:
输入 : L = -10，R = 10，K = 8 输出 : -5 说明:范围内的奇数为-9、-7、-5、-3、-1、3、5、7、9，第八大奇数为-5
输入 : L = -3，R = 3，K = 1 T3】输出 : 3
逼近:给定的问题可以用数学求解。想法是检查 R 是奇数还是偶数，并据此计算第 k 个最大奇数。以下步骤可用于解决问题:
- 如果 K < =0 ，则返回 0
- 初始化计数计算范围内的奇数个数
- 如果 R 是奇数:
- 计数 = ceil((浮动)(R-L+1)/2)
- 如果KT2>计数返回 0
- 否则返回(R2 * K+2)
- 如果 R 是甚至
- 计数 =楼层((R-L+1)/2)
- 如果KT2>计数返回 0
- 否则返回(R2 * K+1)
下面是上述方法的实现:
C++
// C++ implementation for the above approach
#include <cmath>
#include <iostream>
using namespace std;
// Function to return Kth largest
// odd number if it exists
int kthOdd(pair<int, int> 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<int, int> p = { -10, 10 };
// Initialize k
int k = 8;
// print the kth odd number
cout << kthOdd(p, k);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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.
java 描述语言
<script>
// 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
</script>
Output
-5
时间 复杂度:O(1) T5】辅助 空间 : O(1)
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