给定一个n大小的未排序数组，请使用计数排序技术找到中位数和众数。当数组元素在有限范围内时，Thia很有用。
例子：
Input : array a[] = {1, 1, 1, 2, 7, 1}
Output : Mode = 1
Median = 1
Note : Median is average of middle two numbers (1 and 1)
Input : array a[] = {9, 9, 9, 9, 9}
Output : Mode = 9
Median = 9
先决条件：计数排序，数组中位数，模式(数组中最常出现的元素)
输入= [1、4、1、2、7、1、2、5、3、6]
1.辅助(计数)数组在对之前的计数c []求和之前：
索引：0 1 2 3 4 5 6 7 8 9 10
计数：0 3 2 1 1 1 1 1 0 0 0
2.模式=具有最大计数值的索引。
模式= 1(以上示例)
3.计数数组的修改与执行计数排序时所做的修改类似。
索引：0 1 2 3 4 5 6 7 8 9 10
计数：0 3 5 6 7 8 9 10 10 10 10
4.输出数组通常按计数排序b []计算：
输出数组b [] = {1、1、1、2、2、3、4、5、6、7}
5.如果数组b []的大小为奇数，则Median = b [n / 2]
其他中位数=(b [(n-1)/ 2] + b [n / 2])/ 2
6.对于上述示例，b []的大小因此甚至是，Median =(b [4] + b [5])/ 2。
中位数=(2 + 3)/ 2 = 2.5
要遵循的基本方法：
假设输入数组的大小为n ：
步骤1：在将其先前的计数求和成下一个索引之前，先获取计数数组。
步骤2：其中存储有最大值的索引是给定数据的模式。
步骤3：如果其中有多个具有最大值的索引，则所有都是mode的结果，因此我们可以取任何一个。
第4步：将值存储在该索引的一个单独的变量中，该变量称为mode。
步骤5：继续进行计数排序的常规处理。
步骤6：在结果数组(排序后)中，如果n为奇数，则中位数=元素的最中间元素
排序的数组，如果n甚至是中位数=排序数组的两个最中间元素的平均值。
步骤7：将结果存储在称为中值的单独变量中。
以下是上面讨论的问题的实现：
C++
// C++ Program for Mode and
// Median using Counting
// Sort technique
#include
using namespace std;
// function that sort input array a[] and
// calculate mode and median using counting
// sort.
void printModeMedian(int a[], int n)
{
// The output array b[] will
// have sorted array
int b[n];
// variable to store max of
// input array which will
// to have size of count array
int max = *max_element(a, a+n);
// auxiliary(count) array to
// store count. Initialize
// count array as 0. Size
// of count array will be
// equal to (max + 1).
int t = max + 1;
int count[t];
for (int i = 0; i < t; i++)
count[i] = 0;
// Store count of each element
// of input array
for (int i = 0; i < n; i++)
count[a[i]]++;
// mode is the index with maximum count
int mode = 0;
int k = count[0];
for (int i = 1; i < t; i++)
{
if (count[i] > k)
{
k = count[i];
mode = i;
}
}
// Update count[] array with sum
for (int i = 1; i < t; i++)
count[i] = count[i] + count[i-1];
// Sorted output array b[]
// to calculate median
for (int i = 0; i < n; i++)
{
b[count[a[i]]-1] = a[i];
count[a[i]]--;
}
// Median according to odd and even
// array size respectively.
float median;
if (n % 2 != 0)
median = b[n/2];
else
median = (b[(n-1)/2] +
b[(n/2)])/2.0;
// Output the result
cout << "median = " << median << endl;
cout << "mode = " << mode;
}
// Driver program
int main()
{
int a[] = { 1, 4, 1, 2, 7, 1, 2, 5, 3, 6 };
int n = sizeof(a)/sizeof(a[0]);
printModeMedian(a, n);
return 0;
} Java
import java.util.Arrays;
// Java Program for Mode and
// Median using Counting
// Sort technique
class GFG
{
// function that sort input array a[] and
// calculate mode and median using counting
// sort.
static void printModeMedian(int a[], int n)
{
// The output array b[] will
// have sorted array
int[] b = new int[n];
// variable to store max of
// input array which will
// to have size of count array
int max = Arrays.stream(a).max().getAsInt();
// auxiliary(count) array to
// store count. Initialize
// count array as 0. Size
// of count array will be
// equal to (max + 1).
int t = max + 1;
int count[] = new int[t];
for (int i = 0; i < t; i++)
{
count[i] = 0;
}
// Store count of each element
// of input array
for (int i = 0; i < n; i++)
{
count[a[i]]++;
}
// mode is the index with maximum count
int mode = 0;
int k = count[0];
for (int i = 1; i < t; i++)
{
if (count[i] > k)
{
k = count[i];
mode = i;
}
}
// Update count[] array with sum
for (int i = 1; i < t; i++)
{
count[i] = count[i] + count[i - 1];
}
// Sorted output array b[]
// to calculate median
for (int i = 0; i < n; i++)
{
b[count[a[i]] - 1] = a[i];
count[a[i]]--;
}
// Median according to odd and even
// array size respectively.
float median;
if (n % 2 != 0)
{
median = b[n / 2];
}
else
{
median = (float) ((b[(n - 1) / 2]
+ b[(n / 2)]) / 2.0);
}
// Output the result
System.out.println("median = " + median);
System.out.println("mode = " + mode);
}
// Driver program
public static void main(String[] args)
{
int a[] = {1, 4, 1, 2, 7, 1, 2, 5, 3, 6};
int n = a.length;
printModeMedian(a, n);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program for Mode and Median
# using Counting Sort technique
# Function that sort input array a[] and
# calculate mode and median using counting sort
def printModeMedian(a, n):
# The output array b[] will
# have sorted array
b = [0] * n
# Variable to store max of input array
# which will to have size of count array
Max = max(a)
# Auxiliary(count) array to store count.
# Initialize count array as 0. Size of
# count array will be equal to (max + 1).
t = Max + 1
count = [0] * t
# Store count of each element
# of input array
for i in range(n):
count[a[i]] += 1
# Mode is the index with maximum count
mode = 0
k = count[0]
for i in range(1, t):
if (count[i] > k):
k = count[i]
mode = i
# Update count[] array with sum
for i in range(1, t):
count[i] = count[i] + count[i - 1]
# Sorted output array b[]
# to calculate median
for i in range(n):
b[count[a[i]] - 1] = a[i]
count[a[i]] -= 1
# Median according to odd and even
# array size respectively.
median = 0.0
if (n % 2 != 0):
median = b[n // 2]
else:
median = ((b[(n - 1) // 2] +
b[n // 2]) / 2.0)
# Output the result
print("median =", median)
print("mode =", mode)
# Driver Code
if __name__ == '__main__':
arr = [ 1, 4, 1, 2, 7, 1, 2, 5, 3, 6]
n = len(arr)
printModeMedian(arr, n)
# This code is contributed by himanshu77C#
// C# Program for Mode and
// Median using Counting
// Sort technique
using System;
using System.Linq;
class GFG
{
// function that sort input array a[] and
// calculate mode and median using counting
// sort.
static void printModeMedian(int []a, int n)
{
// The output array b[] will
// have sorted array
int[] b = new int[n];
// variable to store max of
// input array which will
// to have size of count array
int max = a.Max();
// auxiliary(count) array to
// store count. Initialize
// count array as 0. Size
// of count array will be
// equal to (max + 1).
int t = max + 1;
int []count = new int[t];
for (int i = 0; i < t; i++)
{
count[i] = 0;
}
// Store count of each element
// of input array
for (int i = 0; i < n; i++)
{
count[a[i]]++;
}
// mode is the index with maximum count
int mode = 0;
int k = count[0];
for (int i = 1; i < t; i++)
{
if (count[i] > k)
{
k = count[i];
mode = i;
}
}
// Update count[] array with sum
for (int i = 1; i < t; i++)
{
count[i] = count[i] + count[i - 1];
}
// Sorted output array b[]
// to calculate median
for (int i = 0; i < n; i++)
{
b[count[a[i]] - 1] = a[i];
count[a[i]]--;
}
// Median according to odd and even
// array size respectively.
float median;
if (n % 2 != 0)
{
median = b[n / 2];
}
else
{
median = (float) ((b[(n - 1) / 2] +
b[(n / 2)]) / 2.0);
}
// Output the result
Console.WriteLine("median = " + median);
Console.WriteLine("mode = " + mode);
}
// Driver Code
public static void Main(String[] args)
{
int []a = {1, 4, 1, 2, 7, 1, 2, 5, 3, 6};
int n = a.Length;
printModeMedian(a, n);
}
}
// This code is contributed by Rajput-JiPHP
$k)
{
$k = $count[$i];
$mode = $i;
}
}
// Update count[] array with sum
for ($i = 1; $i < $t; $i++)
$count[$i] = $count[$i] + $count[$i - 1];
// Sorted output array b[]
// to calculate median
for ($i = 0; $i < $n; $i++)
{
$b[$count[$a[$i]] - 1] = $a[$i];
$count[$a[$i]]--;
}
// Median according to odd and even
// array size respectively.
$median;
if ($n % 2 != 0)
$median = $b[$n / 2];
else
$median = ($b[($n - 1) / 2] +
$b[($n / 2)]) / 2.0;
// Output the result
echo "median = ", $median, "\n" ;
echo "mode = " , $mode;
}
// Driver Code
$a = array( 1, 4, 1, 2, 7,
1, 2, 5, 3, 6 );
$n = sizeof($a);
printModeMedian($a, $n);
// This code is contributed by jit_t
?>- 输出：
median = 2.5
mode = 1
时间复杂度= O(N + P)，其中N是输入数组的时间，P是计数数组的时间。
空间复杂度= O(P)，其中P是辅助数组的大小。