数组中的第 Kth 个奇数
原文:https://www.geeksforgeeks.org/kth-odd-number-in-an-array/
给定一个数组 arr[] 和一个整数 K ，任务是从给定的数组中找到 K 第 个奇数元素。 举例:
输入: arr[] = {1，2，3，4，5}，K = 2 输出: 3 3 是来自给定数组 的 2 nd 奇数元素输入: arr[] = {2，4，6，18}，K = 5 输出: -1 给定数组中没有奇数元素。
方法:逐元素遍历数组元素，对于遇到的每个奇数元素，将值 k 递减 1 。如果 k 的值等于 0 ，则打印当前元素。否则遍历完整数组后，如果 k 的值为 > 0 ，则打印 -1 ，因为数组中奇数元素的总数为 < k 。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the kth odd element
// from the array
int kthOdd(int arr[], int n, int k)
{
// Traverse the array
for (int i = 0; i <= n; i++)
{
// If current element is odd
if ((arr[i] % 2) == 1)
k--;
// If kth odd element is found
if (k == 0)
return arr[i];
}
// Total odd elements in the array are < k
return -1;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr)/sizeof(arr[0]);
int k = 2;
cout << (kthOdd(arr, n, k));
return 0;
}
// This code is contributed by jit_t
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
public class GFG {
// Function to return the kth odd element
// from the array
static int kthOdd(int arr[], int n, int k)
{
// Traverse the array
for (int i = 0; i < n; i++) {
// If current element is odd
if (arr[i] % 2 == 1)
k--;
// If kth odd element is found
if (k == 0)
return arr[i];
}
// Total odd elements in the array are < k
return -1;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = arr.length;
int k = 2;
System.out.print(kthOdd(arr, n, k));
}
}
Python 3
# Python3 implementation of the approach
# Function to return the kth odd
# element from the array
def kthOdd (arr, n, k):
# Traverse the array
for i in range(n):
# If current element is odd
if (arr[i] % 2 == 1):
k -= 1;
# If kth odd element is found
if (k == 0):
return arr[i];
# Total odd elements in the
# array are < k
return -1;
# Driver code
arr = [ 1, 2, 3, 4, 5 ];
n = len(arr);
k = 2;
print(kthOdd(arr, n, k));
# This code is contributed by mits
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the kth odd element
// from the array
static int kthOdd(int []arr, int n, int k)
{
// Traverse the array
for (int i = 0; i < n; i++)
{
// If current element is odd
if (arr[i] % 2 == 1)
k--;
// If kth odd element is found
if (k == 0)
return arr[i];
}
// Total odd elements in the array are < k
return -1;
}
// Driver code
public static void Main()
{
int []arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
int k = 2;
Console.WriteLine(kthOdd(arr, n, k));
}
}
// This code is contributed by SoM15242
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the kth odd
// element from the array
function kthOdd ($arr, $n, $k)
{
// Traverse the array
for ($i = 0; $i < $n; $i++)
{
// If current element is odd
if ($arr[$i] % 2 == 1)
$k--;
// If kth odd element is found
if ($k == 0)
return $arr[$i];
}
// Total odd elements in the
// array are < k
return -1;
}
// Driver code
$arr = array( 1, 2, 3, 4, 5 );
$n = sizeof($arr);
$k = 2;
echo (kthOdd($arr, $n, $k));
// This code is contributed by ajit..
?>
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the kth odd element
// from the array
function kthOdd(arr , n , k) {
// Traverse the array
for (i = 0; i < n; i++) {
// If current element is odd
if (arr[i] % 2 == 1)
k--;
// If kth odd element is found
if (k == 0)
return arr[i];
}
// Total odd elements in the array are < k
return -1;
}
// Driver code
var arr = [ 1, 2, 3, 4, 5 ];
var n = arr.length;
var k = 2;
document.write(kthOdd(arr, n, k));
// This code contributed by Rajput-Ji
</script>
Output:
3
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