阵列中异常的数量
给定一个由 N 个整数组成的数组。异常是一个数，它和数组中其他数的绝对差大于 K，其中 K 是给定的正整数。找出异常的数量。
示例:
Input : arr[] = {1, 3, 5}, k = 1
Output : 3
Explanation:
All the numbers in the array are anamolies because
For the number 1 abs(1-3)=2, abs(1-5)=4 which all are greater than 1,
For the number 3 abs(3-1)=2, abs(3-5)=2 which all are again greater than 1
For the number 5 abs(5-1)=4, abs(5-3)=2 which all are again greater than 1
So there are 3 anamolies.
Input : arr[] = {7, 1, 8}, k = 5
Output : 1
简单方法: 我们只需检查每个数字是否满足给定的条件，即与其他每个数字的绝对差值是否大于 K。
C++
// A simple C++ solution to count anomalies in
// an array.
#include<iostream>
using namespace std;
int countAnomalies(int arr[], int n, int k)
{
int res = 0;
for (int i=0; i<n; i++)
{
int j;
for (j=0; j<n; j++)
if (i != j && abs(arr[i] - arr[j]) <= k)
break;
if (j == n)
res++;
}
return res;
}
int main()
{
int arr[] = {7, 1, 8}, k = 5;
int n = sizeof(arr)/sizeof(arr[0]);
cout << countAnomalies(arr, n, k);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// A simple java solution to count
// anomalies in an array.
class GFG
{
static int countAnomalies(int arr[],
int n, int k)
{
int res = 0;
for (int i = 0; i < n; i++)
{
int j;
for (j = 0; j < n; j++)
if (i != j && Math.abs(arr[i] -
arr[j]) <= k)
break;
if (j == n)
res++;
}
return res;
}
// Driver code
public static void main(String args[])
{
int arr[] = {7, 1, 8}, k = 5;
int n = arr.length;
System.out.println(countAnomalies(arr, n, k));
}
}
// This code is contributed by ANKITRAI1
Python 3
# A simple Python3 solution to
# count anomalies in an array.
def countAnomalies(arr, n, k):
res = 0
for i in range(0, n):
j = 0
while j < n:
if i != j and abs(arr[i] - arr[j]) <= k:
break
j += 1
if j == n:
res += 1
return res
# Driver Code
if __name__ == "__main__":
arr = [7, 1, 8]
k = 5
n = len(arr)
print(countAnomalies(arr, n, k))
# This code is contributed by Rituraj Jain
C
// A simple C# solution to count
// anomalies in an array.
using System;
class GFG
{
static int countAnomalies(int[] arr,
int n, int k)
{
int res = 0;
for (int i = 0; i < n; i++)
{
int j;
for (j = 0; j < n; j++)
if (i != j && Math.Abs(arr[i] -
arr[j]) <= k)
break;
if (j == n)
res++;
}
return res;
}
// Driver code
public static void Main()
{
int[] arr = {7, 1, 8};
int k = 5;
int n = arr.Length;
Console.WriteLine(countAnomalies(arr, n, k));
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// A simple PHP solution to count
// anomalies in an array.
function countAnomalies(&$arr, $n, $k)
{
$res = 0;
for ($i = 0; $i < $n; $i++)
{
for ($j = 0; $j < $n; $j++)
if ($i != $j && abs($arr[$i] -
$arr[$j]) <= $k)
break;
if ($j == $n)
$res++;
}
return $res;
}
// Driver Code
$arr = array(7, 1, 8);
$k = 5;
$n = sizeof($arr);
echo countAnomalies($arr, $n, $k);
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// A simple javascript solution to count
// anomalies in an array.
function countAnomalies(arr,n,k)
{
let res = 0;
for (let i = 0; i < n; i++)
{
let j;
for (j = 0; j < n; j++)
if (i != j && Math.abs(arr[i] -
arr[j]) <= k)
break;
if (j == n)
res++;
}
return res;
}
// Driver code
let arr=[7, 1, 8];
let k = 5;
let n = arr.length;
document.write(countAnomalies(arr, n, k));
// This code is contributed by avanitrachhadiya2155
</script>
Output:
1
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