# 求两个大小相同的排序数组的中位数

2021年3月10日15:52:19 发表评论 538 次浏览

## C

``````// A Simple Merge based O(n) solution to find median of
// two sorted arrays
#include <stdio.h>

/* This function returns median of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[] are sorted arrays
Both have n elements */
int getMedian( int ar1[], int ar2[], int n)
{
int i = 0;  /* Current index of i/p array ar1[] */
int j = 0; /* Current index of i/p array ar2[] */
int count;
int m1 = -1, m2 = -1;

/* Since there are 2n elements, median will be average
of elements at index n-1 and n in the array obtained after
merging ar1 and ar2 */
for (count = 0; count <= n; count++)
{
/*Below is to handle case where all elements of ar1[] are
smaller than smallest(or first) element of ar2[]*/
if (i == n)
{
m1 = m2;
m2 = ar2;
break ;
}

/*Below is to handle case where all elements of ar2[] are
smaller than smallest(or first) element of ar1[]*/
else if (j == n)
{
m1 = m2;
m2 = ar1;
break ;
}
/* equals sign because if two
arrays have some common elements */
if (ar1[i] <= ar2[j])
{
m1 = m2;  /* Store the prev median */
m2 = ar1[i];
i++;
}
else
{
m1 = m2;  /* Store the prev median */
m2 = ar2[j];
j++;
}
}

return (m1 + m2)/2;
}

/* Driver program to test above function */
int main()
{
int ar1[] = {1, 12, 15, 26, 38};
int ar2[] = {2, 13, 17, 30, 45};

int n1 = sizeof (ar1)/ sizeof (ar1);
int n2 = sizeof (ar2)/ sizeof (ar2);
if (n1 == n2)
printf ( "Median is %d" , getMedian(ar1, ar2, n1));
else
printf ( "Doesn't work for arrays of unequal size" );
getchar ();
return 0;
}``````

## C ++

``````// A Simple Merge based O(n)
// solution to find median of
// two sorted arrays
#include <bits/stdc++.h>
using namespace std;

/* This function returns
median of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[]
are sorted arrays
Both have n elements */
int getMedian( int ar1[], int ar2[], int n)
{
int i = 0; /* Current index of
i/p array ar1[] */
int j = 0; /* Current index of
i/p array ar2[] */
int count;
int m1 = -1, m2 = -1;

/* Since there are 2n elements, median will be average of elements
at index n-1 and n in the array
obtained after merging ar1 and ar2 */
for (count = 0; count <= n; count++)
{
/* Below is to handle case where
all elements of ar1[] are
smaller than smallest(or first)
element of ar2[]*/
if (i == n)
{
m1 = m2;
m2 = ar2;
break ;
}

/*Below is to handle case where
all elements of ar2[] are
smaller than smallest(or first)
element of ar1[]*/
else if (j == n)
{
m1 = m2;
m2 = ar1;
break ;
}
/* equals sign because if two
arrays have some common elements */
if (ar1[i] <= ar2[j])
{
/* Store the prev median */
m1 = m2;
m2 = ar1[i];
i++;
}
else
{
/* Store the prev median */
m1 = m2;
m2 = ar2[j];
j++;
}
}

return (m1 + m2)/2;
}

// Driver Code
int main()
{
int ar1[] = {1, 12, 15, 26, 38};
int ar2[] = {2, 13, 17, 30, 45};

int n1 = sizeof (ar1) / sizeof (ar1);
int n2 = sizeof (ar2) / sizeof (ar2);
if (n1 == n2)
cout << "Median is "
<< getMedian(ar1, ar2, n1) ;
else
cout << "Doesn't work for arrays"
<< " of unequal size" ;
getchar ();
return 0;
}

// This code is contributed
// by Shivi_Aggarwal``````

## Java

``````// A Simple Merge based O(n) solution
// to find median of two sorted arrays

class Main
{
// function to calculate median
static int getMedian( int ar1[], int ar2[], int n)
{
int i = 0 ;
int j = 0 ;
int count;
int m1 = - 1 , m2 = - 1 ;

/* Since there are 2n elements, median will
be average of elements at index n-1 and
n in the array obtained after merging ar1
and ar2 */
for (count = 0 ; count <= n; count++)
{
/* Below is to handle case where all
elements of ar1[] are smaller than
smallest(or first) element of ar2[] */
if (i == n)
{
m1 = m2;
m2 = ar2[ 0 ];
break ;
}

/* Below is to handle case where all
elements of ar2[] are smaller than
smallest(or first) element of ar1[] */
else if (j == n)
{
m1 = m2;
m2 = ar1[ 0 ];
break ;
}
/* equals sign because if two
arrays have some common elements */
if (ar1[i] <= ar2[j])
{
/* Store the prev median */
m1 = m2;
m2 = ar1[i];
i++;
}
else
{
/* Store the prev median */
m1 = m2;
m2 = ar2[j];
j++;
}
}

return (m1 + m2)/ 2 ;
}

/* Driver program to test above function */
public static void main (String[] args)
{
int ar1[] = { 1 , 12 , 15 , 26 , 38 };
int ar2[] = { 2 , 13 , 17 , 30 , 45 };

int n1 = ar1.length;
int n2 = ar2.length;
if (n1 == n2)
System.out.println( "Median is " +
getMedian(ar1, ar2, n1));
else
System.out.println( "arrays are of unequal size" );
}
}``````

## Python3

``````# A Simple Merge based O(n) Python 3 solution
# to find median of two sorted lists

# This function returns median of ar1[] and ar2[].
# Assumptions in this function:
# Both ar1[] and ar2[] are sorted arrays
# Both have n elements
def getMedian( ar1, ar2 , n):
i = 0 # Current index of i/p list ar1[]

j = 0 # Current index of i/p list ar2[]

m1 = - 1
m2 = - 1

# Since there are 2n elements, median
# will be average of elements at index
# n-1 and n in the array obtained after
# merging ar1 and ar2
count = 0
while count < n + 1 :
count + = 1

# Below is to handle case where all
# elements of ar1[] are smaller than
# smallest(or first) element of ar2[]
if i = = n:
m1 = m2
m2 = ar2[ 0 ]
break

# Below is to handle case where all
# elements of ar2[] are smaller than
# smallest(or first) element of ar1[]
elif j = = n:
m1 = m2
m2 = ar1[ 0 ]
break
# equals sign because if two
# arrays have some common elements
if ar1[i] < = ar2[j]:
m1 = m2 # Store the prev median
m2 = ar1[i]
i + = 1
else :
m1 = m2 # Store the prev median
m2 = ar2[j]
j + = 1
return (m1 + m2) / 2

# Driver code to test above function
ar1 = [ 1 , 12 , 15 , 26 , 38 ]
ar2 = [ 2 , 13 , 17 , 30 , 45 ]
n1 = len (ar1)
n2 = len (ar2)
if n1 = = n2:
print ( "Median is " , getMedian(ar1, ar2, n1))
else :
print ( "Doesn't work for arrays of unequal size" )

# This code is contributed by "Sharad_Bhardwaj".``````

## C#

``````// A Simple Merge based O(n) solution
// to find median of two sorted arrays
using System;
class GFG
{
// function to calculate median
static int getMedian( int []ar1, int []ar2, int n)
{
int i = 0;
int j = 0;
int count;
int m1 = -1, m2 = -1;

// Since there are 2n elements, // median will be average of
// elements at index n-1 and n in
// the array obtained after
// merging ar1 and ar2
for (count = 0; count <= n; count++)
{
// Below is to handle case
// where all elements of ar1[]
// are smaller than smallest
// (or first) element of ar2[]
if (i == n)
{
m1 = m2;
m2 = ar2;
break ;
}

/* Below is to handle case where all
elements of ar2[] are smaller than
smallest(or first) element of ar1[] */
else if (j == n)
{
m1 = m2;
m2 = ar1;
break ;
}
/* equals sign because if two
arrays have some common elements */
if (ar1[i] <= ar2[j])
{
// Store the prev median
m1 = m2;
m2 = ar1[i];
i++;
}
else
{
// Store the prev median
m1 = m2;
m2 = ar2[j];
j++;
}
}

return (m1 + m2)/2;
}

// Driver Code
public static void Main ()
{
int []ar1 = {1, 12, 15, 26, 38};
int []ar2 = {2, 13, 17, 30, 45};

int n1 = ar1.Length;
int n2 = ar2.Length;
if (n1 == n2)
Console.Write( "Median is " +
getMedian(ar1, ar2, n1));
else
Console.Write( "arrays are of unequal size" );
}
}``````

## 的PHP

``````<?php
// A Simple Merge based O(n) solution
// to find median of two sorted arrays

// This function returns median of
// ar1[] and ar2[]. Assumptions in
// this function: Both ar1[] and ar2[]
// are sorted arrays Both have n elements
function getMedian( \$ar1 , \$ar2 , \$n )
{
// Current index of i/p array ar1[]
\$i = 0;

// Current index of i/p array ar2[]
\$j = 0;
\$count ;
\$m1 = -1; \$m2 = -1;

// Since there are 2n elements, // median will be average of elements
// at index n-1 and n in the array
// obtained after merging ar1 and ar2
for ( \$count = 0; \$count <= \$n ; \$count ++)
{
// Below is to handle case where
// all elements of ar1[] are smaller
// than smallest(or first) element of ar2[]
if ( \$i == \$n )
{
\$m1 = \$m2 ;
\$m2 = \$ar2 ;
break ;
}

// Below is to handle case where all
// elements of ar2[] are smaller than
// smallest(or first) element of ar1[]
else if ( \$j == \$n )
{
\$m1 = \$m2 ;
\$m2 = \$ar1 ;
break ;
}
// equals sign because if two
// arrays have some common elements
if ( \$ar1 [ \$i ] <= \$ar2 [ \$j ])
{
// Store the prev median
\$m1 = \$m2 ;
\$m2 = \$ar1 [ \$i ];
\$i ++;
}
else
{
// Store the prev median
\$m1 = \$m2 ;
\$m2 = \$ar2 [ \$j ];
\$j ++;
}
}

return ( \$m1 + \$m2 ) / 2;
}

// Driver Code
\$ar1 = array (1, 12, 15, 26, 38);
\$ar2 = array (2, 13, 17, 30, 45);

\$n1 = sizeof( \$ar1 );
\$n2 = sizeof( \$ar2 );
if ( \$n1 == \$n2 )
echo ( "Median is " .
getMedian( \$ar1 , \$ar2 , \$n1 ));
else
echo ( "Doesn't work for arrays" .
"of unequal size" );

// This code is contributed by Ajit.
?>``````

``Median is 16``

``````1) Calculate the medians m1 and m2 of the input arrays ar1[]
and ar2[] respectively.
2) If m1 and m2 both are equal then we are done.
return m1 (or m2)
3) If m1 is greater than m2, then median is present in one
of the below two subarrays.
a)  From first element of ar1 to m1 (ar1[0...|_n/2_|])
b)  From m2 to last element of ar2  (ar2[|_n/2_|...n-1])
4) If m2 is greater than m1, then median is present in one
of the below two subarrays.
a)  From m1 to last element of ar1  (ar1[|_n/2_|...n-1])
b)  From first element of ar2 to m2 (ar2[0...|_n/2_|])
5) Repeat the above process until size of both the subarrays
becomes 2.
6) If size of the two arrays is 2 then use below formula to get
the median.
Median = (max(ar1, ar2) + min(ar1, ar2))/2``````

``````ar1[] = {1, 12, 15, 26, 38}
ar2[] = {2, 13, 17, 30, 45}``````

``[15, 26, 38] and [2, 13, 17]``

``m1 = 26 m2 = 13.``

m1大于m2。所以子数组变成

``````[15, 26] and [13, 17]
Now size is 2, so median = (max(ar1, ar2) + min(ar1, ar2))/2
= (max(15, 13) + min(26, 17))/2
= (15 + 17)/2
= 16``````

## C

``````// A divide and conquer based efficient solution to find median
// of two sorted arrays of same size.
#include<bits/stdc++.h>
using namespace std;

int median( int [], int ); /* to get median of a sorted array */

/* This function returns median of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[] are sorted arrays
Both have n elements */
int getMedian( int ar1[], int ar2[], int n)
{
/* return -1  for invalid input */
if (n <= 0)
return -1;
if (n == 1)
return (ar1 + ar2)/2;
if (n == 2)
return (max(ar1, ar2) + min(ar1, ar2)) / 2;

int m1 = median(ar1, n); /* get the median of the first array */
int m2 = median(ar2, n); /* get the median of the second array */

/* If medians are equal then return either m1 or m2 */
if (m1 == m2)
return m1;

/* if m1 < m2 then median must exist in ar1[m1....] and
ar2[....m2] */
if (m1 < m2)
{
if (n % 2 == 0)
return getMedian(ar1 + n/2 - 1, ar2, n - n/2 +1);
return getMedian(ar1 + n/2, ar2, n - n/2);
}

/* if m1 > m2 then median must exist in ar1[....m1] and
ar2[m2...] */
if (n % 2 == 0)
return getMedian(ar2 + n/2 - 1, ar1, n - n/2 + 1);
return getMedian(ar2 + n/2, ar1, n - n/2);
}

/* Function to get median of a sorted array */
int median( int arr[], int n)
{
if (n%2 == 0)
return (arr[n/2] + arr[n/2-1])/2;
else
return arr[n/2];
}

/* Driver program to test above function */
int main()
{
int ar1[] = {1, 2, 3, 6};
int ar2[] = {4, 6, 8, 10};
int n1 = sizeof (ar1)/ sizeof (ar1);
int n2 = sizeof (ar2)/ sizeof (ar2);
if (n1 == n2)
printf ( "Median is %d" , getMedian(ar1, ar2, n1));
else
printf ( "Doesn't work for arrays of unequal size" );
return 0;
}``````

## C ++

``````// A divide and conquer based
// efficient solution to find
// median of two sorted arrays
// of same size.
#include<bits/stdc++.h>
using namespace std;

/* to get median of a
sorted array */
int median( int [], int );

/* This function returns median
of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[] are
sorted arrays
Both have n elements */
int getMedian( int ar1[], int ar2[], int n)
{
/* return -1 for
invalid input */
if (n <= 0)
return -1;
if (n == 1)
return (ar1 +
ar2) / 2;
if (n == 2)
return (max(ar1, ar2) +
min(ar1, ar2)) / 2;

/* get the median of
the first array */
int m1 = median(ar1, n);

/* get the median of
the second array */
int m2 = median(ar2, n);

/* If medians are equal then
return either m1 or m2 */
if (m1 == m2)
return m1;

/* if m1 < m2 then median must
exist in ar1[m1....] and
ar2[....m2] */
if (m1 < m2)
{
if (n % 2 == 0)
return getMedian(ar1 + n / 2 - 1, ar2, n - n / 2 + 1);
return getMedian(ar1 + n / 2, ar2, n - n / 2);
}

/* if m1 > m2 then median must
exist in ar1[....m1] and
ar2[m2...] */
if (n % 2 == 0)
return getMedian(ar2 + n / 2 - 1, ar1, n - n / 2 + 1);
return getMedian(ar2 + n / 2, ar1, n - n / 2);
}

/* Function to get median
of a sorted array */
int median( int arr[], int n)
{
if (n % 2 == 0)
return (arr[n / 2] +
arr[n / 2 - 1]) / 2;
else
return arr[n / 2];
}

// Driver code
int main()
{
int ar1[] = {1, 2, 3, 6};
int ar2[] = {4, 6, 8, 10};
int n1 = sizeof (ar1) /
sizeof (ar1);
int n2 = sizeof (ar2) /
sizeof (ar2);
if (n1 == n2)
cout << "Median is "
<< getMedian(ar1, ar2, n1);
else
cout << "Doesn't work for arrays "
<< "of unequal size" ;
return 0;
}

// This code is contributed
// by Shivi_Aggarwal``````

## Java

``````// A Java program to divide and conquer based
// efficient solution to find
// median of two sorted arrays
// of same size.
import java.util.*;
class GfG {

/* This function returns median
of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[] are
sorted arrays
Both have n elements */

static int getMedian(
int [] a, int [] b, int startA, int startB, int endA, int endB)
{
if (endA - startA == 1 ) {
return (
Math.max(a[startA], b[startB])
+ Math.min(a[endA], b[endB]))
/ 2 ;
}
/* get the median of
the first array */
int m1 = median(a, startA, endA);

/* get the median of
the second array */
int m2 = median(b, startB, endB);

/* If medians are equal then
return either m1 or m2 */
if (m1 == m2) {
return m1;
}

/* if m1 < m2 then median must
exist in ar1[m1....] and
ar2[....m2] */
else if (m1 < m2) {
return getMedian(
a, b, (endA + startA + 1 ) / 2 , startB, endA, (endB + startB + 1 ) / 2 );
}

/* if m1 > m2 then median must
exist in ar1[....m1] and
ar2[m2...] */
else {
return getMedian(
a, b, startA, (endB + startB + 1 ) / 2 , (endA + startA + 1 ) / 2 , endB);
}
}

/* Function to get median
of a sorted array */
static int median(
int [] arr, int start, int end)
{
int n = end - start + 1 ;
if (n % 2 == 0 ) {
return (
arr[start + (n / 2 )]
+ arr[start + (n / 2 - 1 )])
/ 2 ;
}
else {
return arr[start + n / 2 ];
}
}

// Driver code
public static void main(String[] args)
{
int ar1[] = { 1 , 2 , 3 , 6 };
int ar2[] = { 4 , 6 , 8 , 10 };
int n1 = ar1.length;
int n2 = ar2.length;
if (n1 != n2) {
System.out.println(
"Doesn't work for arrays "
+ "of unequal size" );
}
else if (n1 == 0 ) {
System.out.println( "Arrays are empty." );
}
else if (n1 == 1 ) {
System.out.println((ar1[ 0 ] + ar2[ 0 ]) / 2 );
}
else {
System.out.println(
"Median is "
+ getMedian(
ar1, ar2, 0 , 0 , ar1.length - 1 , ar2.length - 1 ));
}
}
}``````

## python

``````# using divide and conquer we divide
# the 2 arrays accordingly recursively
# till we get two elements in each
# array, hence then we calculate median

#condition len(arr1)=len(arr2)=n
def getMedian(arr1, arr2, n):

# there is no element in any array
if n = = 0 :
return - 1

# 1 element in each => median of
# sorted arr made of two arrays will
elif n = = 1 :
# be sum of both elements by 2
return (arr1[ 0 ] + arr2[ 1 ]) / 2

# Eg. [1, 4] , [6, 10] => [1, 4, 6, 10]
# median = (6+4)/2
elif n = = 2 :
# which implies median = (max(arr1, # arr2)+min(arr1, arr2))/2
return ( max (arr1[ 0 ], arr2[ 0 ]) +
min (arr1[ 1 ], arr2[ 1 ])) / 2

else :
#calculating medians
m1 = median(arr1, n)
m2 = median(arr2, n)

# then the elements at median
# position must be between the
# greater median and the first
# element of respective array and
# between the other median and
# the last element in its respective array.
if m1 > m2:

if n % 2 = = 0 :
return getMedian(arr1[: int (n / 2 ) + 1 ], arr2[ int (n / 2 ) - 1 :], int (n / 2 ) + 1 )
else :
return getMedian(arr1[: int (n / 2 ) + 1 ], arr2[ int (n / 2 ):], int (n / 2 ) + 1 )

else :
if n % 2 = = 0 :
return getMedian(arr1[ int (n / 2 - 1 ):], arr2[: int (n / 2 + 1 )], int (n / 2 ) + 1 )
else :
return getMedian(arr1[ int (n / 2 ):], arr2[ 0 : int (n / 2 ) + 1 ], int (n / 2 ) + 1 )

# function to find median of array
def median(arr, n):
if n % 2 = = 0 :
return (arr[ int (n / 2 )] +
arr[ int (n / 2 ) - 1 ]) / 2
else :
return arr[ int (n / 2 )]

# Driver code
arr1 = [ 1 , 2 , 3 , 6 ]
arr2 = [ 4 , 6 , 8 , 10 ]
n = len (arr1)
print ( int (getMedian(arr1, arr2, n)))

# This code is contributed by
# baby_gog9800``````

## C#

``````// A C# program to divide and conquer based
// efficient solution to find
// median of two sorted arrays
// of same size.
using System;
class GfG{

/* This function returns median
of ar1[] and ar2[].
Assumptions in this function:
Both ar1[] and ar2[] are
sorted arrays
Both have n elements */

static int getMedian( int [] a, int [] b, int startA, int startB, int endA, int endB)
{
if (endA - startA == 1)
{
return (Math.Max(a[startA], b[startB]) +
Math.Min(a[endA], b[endB])) / 2;
}

/* get the median of
the first array */
int m1 = median(a, startA, endA);

/* get the median of
the second array */
int m2 = median(b, startB, endB);

/* If medians are equal then
return either m1 or m2 */
if (m1 == m2)
{
return m1;
}

/*if m1 < m2 then median must
exist in ar1[m1....] and
ar2[....m2] */
else if (m1 < m2)
{
return getMedian(a, b, (endA + startA + 1) / 2, startB, endA, (endB + startB + 1) / 2);
}

/*if m1 > m2 then median must
exist in ar1[....m1] and
ar2[m2...] */
else
{
return getMedian(a, b, startA, (endB + startB + 1) / 2, (endA + startA + 1) / 2, endB);
}
}

/* Function to get median
of a sorted array */
static int median( int [] arr, int start, int end)
{
int n = end - start + 1;
if (n % 2 == 0)
{
return (arr[start + (n / 2)] +
arr[start + (n / 2 - 1)]) / 2;
}
else
{
return arr[start + n / 2];
}
}

// Driver code
public static void Main(String[] args)
{
int []ar1 = {1, 2, 3, 6};
int []ar2 = {4, 6, 8, 10};
int n1 = ar1.Length;
int n2 = ar2.Length;
if (n1 != n2)
{
Console.WriteLine( "Doesn't work for arrays " +
"of unequal size" );
}
else if (n1 == 0)
{
Console.WriteLine( "Arrays are empty." );
}
else if (n1 == 1)
{
Console.WriteLine((ar1 + ar2) / 2);
}
else
{
Console.WriteLine( "Median is " +
getMedian(ar1, ar2, 0, 0, ar1.Length - 1, ar2.Length - 1));
}
}
}

// This code is contributed by gauravrajput1``````

``Median is 5``

O(登录)

http://en.wikipedia.org/wiki/Median

http://ocw.alfaisal.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-Science/6-046JFall-2005/30C68118-E436-4FE3-8C79-6BAFBB07D935/0/ps9sol.pdf ds3etph5wn 