给定n个元素的未排序数组，并给定两个点num1和num2。任务是计算给定点之间(不包括num1和num2)之间发生的元素数量。
如果存在num1和num2多次出现，则需要考虑最左边的num1和最右边的num2。
例子：
Input : arr[] = {3 5 7 6 4 9 12 4 8}
num1 = 5
num2 = 4
Output : 5
Number of elements between leftmost occurrence
of 5 and rightmost occurrence of 4 is five.
Input : arr[] = {4, 6, 8, 3, 6, 2, 8, 9, 4}
num1 = 4
num2 = 4
Output : 7
Input : arr[] = {4, 6, 8, 3, 6, 2, 8, 9, 4}
num1 = 4
num2 = 10
Output : 0在所有情况下(当不存在单个或两个元素时)，解决方案应仅遍历数组一次
这个想法是从左边遍历数组并找到第一个出现的num1。如果到达末尾，则返回0。然后从最右边的元素开始遍历并找到num2。我们仅遍历直到大于num1的索引的点。如果到达末尾，则返回0。如果找到两个元素，则使用找到的元素的索引返回计数。
CPP
// Program to count number of elements between
// two given elements.
#include
using namespace std;
// Function to count number of elements
// occurs between the elements.
int getCount(int arr[], int n, int num1, int num2)
{
// Find num1
int i = 0;
for (i = 0; i < n; i++)
if (arr[i] == num1)
break;
// If num1 is not present or present at end
if (i >= n-1)
return 0;
// Find num2
int j;
for (j = n-1; j >= i+1; j--)
if (arr[j] == num2)
break;
// If num2 is not present
if (j == i)
return 0;
// return number of elements between
// the two elements.
return (j - i - 1);
}
// Driver Code
int main()
{
int arr[] = { 3, 5, 7, 6, 4, 9, 12, 4, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
int num1 = 5, num2 = 4;
cout << getCount(arr, n, num1, num2);
return 0;
} Java
// Program to count number of elements
// between two given elements.
import java.io.*;
class GFG
{
// Function to count number of elements
// occurs between the elements.
static int getCount(int arr[], int n,
int num1, int num2)
{
// Find num1
int i = 0;
for (i = 0; i < n; i++)
if (arr[i] == num1)
break;
// If num1 is not present
// or present at end
if (i >= n - 1)
return 0;
// Find num2
int j;
for (j = n - 1; j >= i + 1; j--)
if (arr[j] == num2)
break;
// If num2 is not present
if (j == i)
return 0;
// return number of elements
// between the two elements.
return (j - i - 1);
}
// Driver program
public static void main (String[] args)
{
int arr[] = { 3, 5, 7, 6, 4, 9, 12, 4, 8 };
int n = arr.length;
int num1 = 5, num2 = 4;
System.out.println( getCount(arr, n, num1, num2));
}
}
// This code is contributed by vt_mPython3
# Python Program to count number of elements between
# two given elements.
# Function to count number of elements
# occurs between the elements.
def getCount(arr, n, num1, num2):
# Find num1
for i in range(0,n):
if (arr[i] == num1):
break
#If num1 is not present or present at end
if (i >= n-1):
return 0
# Find num2
for j in range(n-1, i+1, -1):
if (arr[j] == num2):
break
# If num2 is not present
if (j == i):
return 0
# return number of elements between
# the two elements.
return (j - i - 1)
# Driver Code
arr= [ 3, 5, 7, 6, 4, 9, 12, 4, 8 ]
n=len(arr)
num1 = 5
num2 = 4
print(getCount(arr, n, num1, num2))
# This code is contributed by SHARIQ_JMIC#
// C# Program to count number of elements
// between two given elements.
using System;
class GFG {
// Function to count number of elements
// occurs between the elements.
static int getCount(int []arr, int n,
int num1, int num2)
{
// Find num1
int i = 0;
for (i = 0; i < n; i++)
if (arr[i] == num1)
break;
// If num1 is not present
// or present at end
if (i >= n - 1)
return 0;
// Find num2
int j;
for (j = n - 1; j >= i + 1; j--)
if (arr[j] == num2)
break;
// If num2 is not present
if (j == i)
return 0;
// return number of elements
// between the two elements.
return (j - i - 1);
}
// Driver Code
public static void Main ()
{
int []arr = {3, 5, 7, 6, 4, 9, 12, 4, 8};
int n = arr.Length;
int num1 = 5, num2 = 4;
Console.WriteLine(getCount(arr, n, num1, num2));
}
}
// This article is contributed by vt_m.PHP
= $n - 1)
return 0;
// Find num2
$j;
for ($j = $n - 1; $j >= $i + 1; $j--)
if ($arr[$j] == $num2)
break;
// If num2 is not present
if ($j == $i)
return 0;
// return number of elements
// betweenthe two elements.
return ($j - $i - 1);
}
// Driver Code
$arr = array(3, 5, 7, 6, 4, 9, 12, 4, 8);
$n = sizeof($arr);
$num1 = 5; $num2 = 4;
echo(getCount($arr, $n, $num1, $num2));
// This code is contributed by Ajit.
?>Javascript
输出：
5