两个大小相同的排序数组的中位数

有2个大小为n的排序数组A和B。编写算法以查找合并上述两个数组(即长度为2n的数组)后获得的数组的中位数。复杂度应为O(log(n))。

注意：由于我们要寻找中位数的集合的大小为偶数(2n)，因此我们需要取中间两个数字的平均值，然后返回平均值的下限。
方法1(合并时简单计数)
使用合并排序的合并过程。比较两个数组的元素时，请跟踪计数。如果count变成n(对于2n个元素)，则我们达到了中位数。取合并数组中索引n-1和n处元素的平均值。请参见下面的实现。

C

// A Simple Merge based O(n) solution to find median of
// two sorted arrays
#include 
 
/* 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[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])
        {
            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[0]);
    int n2 = sizeof(ar2)/sizeof(ar2[0]);
    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 
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[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 Code
int main()
{
    int ar1[] = {1, 12, 15, 26, 38};
    int ar2[] = {2, 13, 17, 30, 45};
 
    int n1 = sizeof(ar1) / sizeof(ar1[0]);
    int n2 = sizeof(ar2) / sizeof(ar2[0]);
    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[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 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");
    }
}

Javascript

 
// 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)
{
    var i = 0; /* Current index of i/p array ar1[] */
    var j = 0; /* Current index of i/p array ar2[] */
    var count;
    var 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])
        {
            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 */
var ar1 = [1, 12, 15, 26, 38];
var ar2 = [2, 13, 17, 30, 45];
var n1 = ar1.length;
var n2 = ar2.length;
if (n1 == n2)
    document.write("Median is "+ getMedian(ar1, ar2, n1));
else
    document.write("Doesn't work for arrays of unequal size");
 

输出 ：

Median is 16

时间复杂度： O(n)
方法2(通过比较两个数组的中位数)
此方法的工作原理是首先获取两个排序数组的中值，然后进行比较。
令ar1和ar2为输入数组。
算法 ：

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[0], ar2[0]) + min(ar1[1], ar2[1]))/2

例子 ：

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

对于上述两个数组，m1 = 15和m2 = 17
对于上述ar1 []和ar2 []，m1小于m2。因此，中位数出现在以下两个子数组之一中。

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

让我们对以上两个子数组重复该过程：

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[0], ar2[0]) + min(ar1[1], ar2[1]))/2
                       = (max(15, 13) + min(26, 17))/2 
                       = (15 + 17)/2
                       = 16

执行 ：

C

// A divide and conquer based efficient solution to find median
// of two sorted arrays of same size.
#include
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[0] + ar2[0])/2;
    if (n == 2)
        return (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1])) / 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[0]);
    int n2 = sizeof(ar2)/sizeof(ar2[0]);
    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
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[0] +
                ar2[0]) / 2;
    if (n == 2)
        return (max(ar1[0], ar2[0]) +
                min(ar1[1], ar2[1])) / 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[0]);
    int n2 = sizeof(ar2) /
             sizeof(ar2[0]);
    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[0])/2
         
    # Eg. [1,4] , [6,10] => [1, 4, 6, 10]
    # median = (6+4)/2   
    elif n == 2:
        # which implies median = (max(arr1[0],
        # arr2[0])+min(arr1[1],arr2[1]))/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[0] + ar2[0]) / 2);
  }
  else
  {
    Console.WriteLine("Median is " +
                       getMedian(ar1, ar2, 0, 0,
                                 ar1.Length - 1,
                                 ar2.Length - 1));
  }
}
}
 
// This code is contributed by gauravrajput1

输出 ：

Median is 5

时间复杂度： O(logn)
算法范式：分而治之

方法3(通过占用无多余空间的工会)

此方法的工作原理是在没有多余空间的情况下合并两个数组的并，然后对其进行排序。

算法 ：

1) Take the union of the input arrays ar1[] and ar2[].
2) Sort ar1[] and ar2[] respectively.
3) The median will be the last element of ar1[] + the first
   element of ar2[] divided by 2. [(ar1[n-1] + ar2[0])/2].

执行 ：

C++

// CPP program for the above approach
#include 
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 j = 0;
    int i = n - 1;
    while (ar1[i] > ar2[j] && j < n && i > -1)
        swap(ar1[i--], ar2[j++]);
    sort(ar1, ar1 + n);
    sort(ar2, ar2 + n);
    return (ar1[n - 1] + ar2[0]) / 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[0]);
    int n2 = sizeof(ar2) / sizeof(ar2[0]);
    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 Lakshay

Java

/*package whatever //do not write package name here */
import java.io.*;
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 */
  public static int getMedian(int ar1[],
                              int ar2[], int n)
  {
    int j = 0;
    int i = n - 1;
    while (ar1[i] > ar2[j] && j < n && i > -1)
    {
      int temp = ar1[i];
      ar1[i] = ar2[j];
      ar2[j] = temp;
      i--; j++;
    }
    Arrays.sort(ar1);
    Arrays.sort(ar2);
    return (ar1[n - 1] + ar2[0]) / 2;
  }
 
  // Driver code
  public static void main (String[] args)
  {
    int ar1[] = { 1, 12, 15, 26, 38 };
    int ar2[] = { 2, 13, 17, 30, 45 };
 
    int n1 = 5;
    int n2 = 5;
    if (n1 == n2)
      System.out.println("Median is "+ getMedian(ar1, ar2, n1));
    else
      System.out.println("Doesn't work for arrays of unequal size");
  }
}
 
// This code is contributed by Manu Pathria

Python3

# Python program for above approach
 
# function to return median of the arrays
# both are sorted & of same size
def getMedian(ar1, ar2, n):
    i, j = n - 1, 0
 
    # while loop to swap all smaller numbers to arr1
    while(ar1[i] > ar2[j] and i > -1 and j < n):
        ar1[i], ar2[j] = ar2[j], ar1[i]
        i -= 1
        j += 1
 
    ar1.sort()
    ar2.sort()
 
    return (ar1[-1] + ar2[0]) >> 1
 
 
# Driver program
if __name__ == '__main__':
    ar1 = [1, 12, 15, 26, 38]
    ar2 = [2, 13, 17, 30, 45]
 
    n1, n2 = len(ar1), 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 saitejagampala

C＃

/*package whatever //do not write package name here */
using System;
public 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 */
  public static int getMedian(int []ar1,
                              int []ar2, int n)
  {
    int j = 0;
    int i = n - 1;
    while (ar1[i] > ar2[j] && j < n && i > -1)
    {
      int temp = ar1[i];
      ar1[i] = ar2[j];
      ar2[j] = temp;
      i--; j++;
    }
    Array.Sort(ar1);
    Array.Sort(ar2);
    return (ar1[n - 1] + ar2[0]) / 2;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int []ar1 = { 1, 12, 15, 26, 38 };
    int []ar2 = { 2, 13, 17, 30, 45 };
 
    int n1 = 5;
    int n2 = 5;
    if (n1 == n2)
      Console.WriteLine("Median is "+ getMedian(ar1, ar2, n1));
    else
      Console.WriteLine("Doesn't work for arrays of unequal size");
  }
}
 
// This code is contributed by aashish1995

输出

Median is 16

时间复杂度： O(nlogn)