O(log(min(n(n, m)))中具有不同大小的两个排序数组的中位数
原文： https://www.geeksforgeeks.org/median-two-sorted-arrays-different-sizes-ologminn-m/
给定两个排序数组a[]和b[]，当n是第一个数组中的元素数时，任务是在O(log(min(n(n, m)))中找到这些排序数组的中位数。 m是第二个数组中的元素数。
先决条件： 两个不同大小的排序数组的中位数。
示例：
Input : ar1[] = {-5, 3, 6, 12, 15}
ar2[] = {-12, -10, -6, -3, 4, 10}
The merged array is :
ar3[] = {-12, -10, -6, -5 , -3,
3, 4, 6, 10, 12, 15}
Output : The median is 3.
Input : ar1[] = {2, 3, 5, 8}
ar2[] = {10, 12, 14, 16, 18, 20}
The merged array is :
ar3[] = {2, 3, 5, 8, 10, 12, 14, 16, 18, 20}
if the number of the elements are even,
so there are two middle elements,
take the average between the two :
(10 + 12) / 2 = 11\.
Output : The median is 11.
注意：如果总数为偶数，并且如果我们要返回合并数组中存在的中位数，则可以返回(n + m) / 2或(n + m) / 2 – 1位。 在这种情况下，中位数可以是 10 或 12。
方法：开始将两个数组划分为两半（不是两个部分，但两个分区都应具有相同数量的元素）。 前半部分包含来自第一个和第二个数组的一些第一元素，后半部分包含形成第一个和第二个数组的其余（或最后一个）元素。 由于数组可以具有不同的大小，因此并不意味着要从每个数组中取出每一半。 以下示例阐明了解释。 达到这样的条件：上半部分的每个元素都小于或等于下半部分的每个元素。
如何达到此条件？
偶数情况下的示例。 假设找到了分区。 因为A[]和B[]是两个排序的数组，所以a1小于或等于a2，b2小于或等于b3。 现在，检查a1是否小于或等于b3，以及b2是否小于或等于a2。 如果是这种情况，则意味着上半部分的每个元素都小于或等于下半部分的每个元素，因为a1大于或等于A[]中的每个元素（a0）和b2大于或等于B[]中它之前的每个元素（b1和b0）。 如果总数为偶数，则中位数将为a1，b2的最大值与a2，b3的最小值之间的平均值，但是如果总数为奇数，则中位数将为a2，b2的最大值。 但是，如果不是这两种情况，则有两种选择（以偶数示例为例）：[b] > a2表示b2 > a2或a1 > b3，在数组的右侧搜索，如果a1 > b3，则意味着在数组的左侧搜索，直到找到所需的条件。
为什么上述情况导致中位数？
中位数是数组中(n + 1) / 2的最小元素，这里，中位数是两个数组中(n + m + 1) / 2的最小元素。 如果上半部分中的所有元素都小于（或等于）下半部分中的所有元素，则在总数为奇数的情况下，只需计算上半部分的最后两个元素之间的最大值（a2和b2我们的示例），这将导致我们找到两个数组中的(n + m + 1) / 2个最小元素，即中位数（(7 + 4 + 1) / 2 = 6）。 但是在总数为偶数的情况下，请计算前半部分中最后两个元素的最大值（在我们的示例中为a1和b2）之间的平均值及其在数组中的连续数，即后半部分中前两个元素的最小值（在我们的示例中为a2和b3）。
分区的过程：
要分成两半，请进行分区，以使分区数组A[]的索引加分区数组B[]的索引等于元素总数加一除以 2，即(n + m + 1) / 2（如果元素总数为奇数，则为+1）。
首先，定义两个变量：min_index和max_index，并将min_index初始化为 0，并将max_index初始化为较小数组的长度。 在以下这些示例中，A[]是较小的数组。
要对A[]进行分区，请使用公式(min_index + max_index) / 2并将其插入变量i。 要对B[]进行分区，请使用公式(n + m + 1) / 2 – i并将其插入变量j。
变量i表示要从A[]插入到前半部分的元素数，而j表示要从B[]插入到前半部分的元素数，其余的元素将被插入后半部分。
请看以下示例：
示例 1：
示例 2 （此示例是指返回合并数组中存在的中值的条件）：
以下是上述方法的实现：
C++
// CPP code for median with case of returning
// double value when even number of elements are
// present in both array combinely
#include<bits/stdc++.h>
using std::cout;
int maximum(int a, int b);
int minimum(int a, int b);
// Function to find median of two sorted arrays
double findMedianSortedArrays(int *a, int n,
int *b, int m)
{
int min_index = 0, max_index = n, i, j, median;
while (min_index <= max_index)
{
i = (min_index + max_index) / 2;
j = ((n + m + 1) / 2) - i;
// If j is negative then the partition is not
// possible having i elements from array i
if (j < 0)
{
max_index = i-1;
continue;
}
// if i = n, it means that Elements from a[] in
// the second half is an empty set. and if j = 0,
// it means that Elements from b[] in the first
// half is an empty set. so it is necessary to
// check that, because we compare elements from
// these two groups.
// Searching on right
if (i < n && j > 0 && b[j - 1] > a[i])
min_index = i + 1;
// if i = 0, it means that Elements from a[] in
// the first half is an empty set and if j = m,
// it means that Elements from b[] in the second
// half is an empty set. so it is necessary to
// check that, because we compare elements
// from these two groups.
// searching on left
else if (i > 0 && j < m && b[j] < a[i - 1])
max_index = i - 1;
// we have found the desired halves.
else
{
// this condition happens when we don't have any
// elements in the first half from a[] so we
// returning the last element in b[] from
// the first half.
if (i == 0)
median = b[j - 1];
// and this condition happens when we don't
// have any elements in the first half from
// b[] so we returning the last element in
// a[] from the first half.
else if (j == 0)
median = a[i - 1];
else
median = maximum(a[i - 1], b[j - 1]);
break;
}
}
// calculating the median.
// If number of elements is odd there is
// one middle element.
if ((n + m) % 2 == 1)
return (double)median;
// Elements from a[] in the second half is an empty set.
if (i == n)
return (median+b[j]) / 2.0;
// Elements from b[] in the second half is an empty set.
if (j == m)
return (median + a[i]) / 2.0;
return (median + minimum(a[i], b[j])) / 2.0;
}
// Function to find max
int maximum(int a, int b)
{
return a > b ? a : b;
}
// Function to find minimum
int minimum(int a, int b)
{
return a < b ? a : b;
}
// Driver code
int main()
{
int a[] = {900};
int b[] = { 10, 13, 14};
int n = sizeof(a) / sizeof(int);
int m = sizeof(b) / sizeof(int);
// we need to define the smaller array as the
// first parameter to make sure that the
// time complexity will be O(log(min(n,m)))
if (n < m)
cout << "The median is : "
<< findMedianSortedArrays(a, n, b, m);
else
cout << "The median is : "
<< findMedianSortedArrays(b, m, a, n);
return 0;
}
Java
// Java code for median with
// case of returning double
// value when even number of
// elements are present in
// both array combinely
import java.io.*;
class GFG
{
static int []a = new int[]{900};
static int []b = new int[]{10, 13, 14};
// Function to find max
static int maximum(int a, int b)
{
return a > b ? a : b;
}
// Function to find minimum
static int minimum(int a, int b)
{
return a < b ? a : b;
}
// Function to find median
// of two sorted arrays
static double findMedianSortedArrays(int n,
int m)
{
int min_index = 0,
max_index = n, i = 0,
j = 0, median = 0;
while (min_index <= max_index)
{
i = (min_index + max_index) / 2;
j = ((n + m + 1) / 2) - i;
// if i = n, it means that Elements
// from a[] in the second half is an
// empty set. and if j = 0, it means
// that Elements from b[] in the first
// half is an empty set. so it is
// necessary to check that, because we
// compare elements from these two
// groups. Searching on right
if (i < n && j > 0 && b[j - 1] > a[i])
min_index = i + 1;
// if i = 0, it means that Elements
// from a[] in the first half is an
// empty set and if j = m, it means
// that Elements from b[] in the second
// half is an empty set. so it is
// necessary to check that, because
// we compare elements from these two
// groups. searching on left
else if (i > 0 && j < m && b[j] < a[i - 1])
max_index = i - 1;
// we have found the desired halves.
else
{
// this condition happens when we
// don't have any elements in the
// first half from a[] so we
// returning the last element in
// b[] from the first half.
if (i == 0)
median = b[j - 1];
// and this condition happens when
// we don't have any elements in the
// first half from b[] so we
// returning the last element in
// a[] from the first half.
else if (j == 0)
median = a[i - 1];
else
median = maximum(a[i - 1],
b[j - 1]);
break;
}
}
// calculating the median.
// If number of elements is odd
// there is one middle element.
if ((n + m) % 2 == 1)
return (double)median;
// Elements from a[] in the
// second half is an empty set.
if (i == n)
return (median + b[j]) / 2.0;
// Elements from b[] in the
// second half is an empty set.
if (j == m)
return (median + a[i]) / 2.0;
return (median + minimum(a[i],
b[j])) / 2.0;
}
// Driver code
public static void main(String args[])
{
int n = a.length;
int m = b.length;
// we need to define the
// smaller array as the
// first parameter to
// make sure that the
// time complexity will
// be O(log(min(n,m)))
if (n < m)
System.out.print("The median is : " +
findMedianSortedArrays(n, m));
else
System.out.print("The median is : " +
findMedianSortedArrays(m, n));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
Python3
# Python code for median with
# case of returning double
# value when even number
# of elements are present
# in both array combinely
median = 0
i = 0
j = 0
# def to find max
def maximum(a, b) :
return a if a > b else b
# def to find minimum
def minimum(a, b) :
return a if a < b else b
# def to find median
# of two sorted arrays
def findMedianSortedArrays(a, n, b, m) :
global median, i, j
min_index = 0
max_index = n
while (min_index <= max_index) :
i = int((min_index + max_index) / 2)
j = int(((n + m + 1) / 2) - i)
# if i = n, it means that
# Elements from a[] in the
# second half is an empty
# set. and if j = 0, it
# means that Elements from
# b[] in the first half is
# an empty set. so it is
# necessary to check that,
# because we compare elements
# from these two groups.
# Searching on right
if (i < n and j > 0 and b[j - 1] > a[i]) :
min_index = i + 1
# if i = 0, it means that
# Elements from a[] in the
# first half is an empty
# set and if j = m, it means
# that Elements from b[] in
# the second half is an empty
# set. so it is necessary to
# check that, because we compare
# elements from these two groups.
# searching on left
elif (i > 0 and j < m and b[j] < a[i - 1]) :
max_index = i - 1
# we have found the
# desired halves.
else :
# this condition happens when
# we don't have any elements
# in the first half from a[]
# so we returning the last
# element in b[] from the
# first half.
if (i == 0) :
median = b[j - 1]
# and this condition happens
# when we don't have any
# elements in the first half
# from b[] so we returning the
# last element in a[] from the
# first half.
elif (j == 0) :
median = a[i - 1]
else :
median = maximum(a[i - 1], b[j - 1])
break
# calculating the median.
# If number of elements
# is odd there is
# one middle element.
if ((n + m) % 2 == 1) :
return median
# Elements from a[] in the
# second half is an empty set.
if (i == n) :
return ((median + b[j]) / 2.0)
# Elements from b[] in the
# second half is an empty set.
if (j == m) :
return ((median + a[i]) / 2.0)
return ((median + minimum(a[i], b[j])) / 2.0)
# Driver code
a = [900]
b = [10, 13, 14]
n = len(a)
m = len(b)
# we need to define the
# smaller array as the
# first parameter to make
# sure that the time complexity
# will be O(log(min(n,m)))
if (n < m) :
print ("The median is : {}".format(findMedianSortedArrays(a, n, b, m)))
else :
echo ("The median is : {}".format(findMedianSortedArrays(b, m, a, n)))
# This code is contributed
# by Manish Shaw(manishshaw1)
C#
// C# code for median with case of returning
// double value when even number of elements
// are present in both array combinely
using System;
class GFG {
// Function to find max
static int maximum(int a, int b)
{
return a > b ? a : b;
}
// Function to find minimum
static int minimum(int a, int b)
{
return a < b ? a : b;
}
// Function to find median of two sorted
// arrays
static double findMedianSortedArrays(ref int []a,
int n, ref int []b, int m)
{
int min_index = 0, max_index = n, i = 0,
j = 0, median = 0;
while (min_index <= max_index)
{
i = (min_index + max_index) / 2;
j = ((n + m + 1) / 2) - i;
// if i = n, it means that Elements
// from a[] in the second half is an
// empty set. and if j = 0, it means
// that Elements from b[] in the first
// half is an empty set. so it is
// necessary to check that, because we
// compare elements from these two
// groups. Searching on right
if (i < n && j > 0 && b[j - 1] > a[i])
min_index = i + 1;
// if i = 0, it means that Elements
// from a[] in the first half is an
// empty set and if j = m, it means
// that Elements from b[] in the second
// half is an empty set. so it is
// necessary to check that, because
// we compare elements from these two
// groups. searching on left
else if (i > 0 && j < m && b[j] < a[i - 1])
max_index = i - 1;
// we have found the desired halves.
else
{
// this condition happens when we
// don't have any elements in the
// first half from a[] so we
// returning the last element in
// b[] from the first half.
if (i == 0)
median = b[j - 1];
// and this condition happens when
// we don't have any elements in the
// first half from b[] so we
// returning the last element in
// a[] from the first half.
else if (j == 0)
median = a[i - 1];
else
median = maximum(a[i - 1], b[j - 1]);
break;
}
}
// calculating the median.
// If number of elements is odd
// there is one middle element.
if ((n + m) % 2 == 1)
return (double)median;
// Elements from a[] in the second
// half is an empty set.
if (i == n)
return (median+b[j]) / 2.0;
// Elements from b[] in the second
// half is an empty set.
if (j == m)
return (median + a[i]) / 2.0;
return (median + minimum(a[i], b[j])) / 2.0;
}
// Driver code
static void Main()
{
int []a = new int[]{900};
int []b = new int[]{ 10, 13, 14};
int n = a.Length;
int m = b.Length;
// we need to define the smaller
// array as the first parameter to
// make sure that the time
// complexity will be O(log(min(n,m)))
if (n < m)
Console.Write("The median is : "
+ findMedianSortedArrays(ref a, n,
ref b, m));
else
Console.Write("The median is : "
+ findMedianSortedArrays(ref b, m,
ref a, n));
}
}
// This code is contributed by Manish Shaw
// (manishshaw1)
PHP
<?php
// PHP code for median with
// case of returning double
// value when even number
// of elements are present
// in both array combinely
$median = 0;
$i = 0; $j = 0;
// Function to find max
function maximum($a, $b)
{
return $a > $b ? $a : $b;
}
// Function to find minimum
function minimum($a, $b)
{
return $a < $b ? $a : $b;
}
// Function to find median
// of two sorted arrays
function findMedianSortedArrays(&$a, $n,
&$b, $m)
{
global $median, $i, $j;
$min_index = 0;
$max_index = $n;
while ($min_index <= $max_index)
{
$i = intval(($min_index +
$max_index) / 2);
$j = intval((($n + $m + 1) /
2) - $i);
// if i = n, it means that
// Elements from a[] in the
// second half is an empty
// set. and if j = 0, it
// means that Elements from
// b[] in the first half is
// an empty set. so it is
// necessary to check that,
// because we compare elements
// from these two groups.
// Searching on right
if ($i < $n && $j > 0 &&
$b[$j - 1] > $a[$i])
$min_index = $i + 1;
// if i = 0, it means that
// Elements from a[] in the
// first half is an empty
// set and if j = m, it means
// that Elements from b[] in
// the second half is an empty
// set. so it is necessary to
// check that, because we compare
// elements from these two groups.
// searching on left
else if ($i > 0 && $j < $m &&
$b[$j] < $a[$i - 1])
$max_index = $i - 1;
// we have found the
// desired halves.
else
{
// this condition happens when
// we don't have any elements
// in the first half from a[]
// so we returning the last
// element in b[] from the
// first half.
if ($i == 0)
$median = $b[$j - 1];
// and this condition happens
// when we don't have any
// elements in the first half
// from b[] so we returning the
// last element in a[] from the
// first half.
else if ($j == 0)
$median = $a[$i - 1];
else
$median = maximum($a[$i - 1],
$b[$j - 1]);
break;
}
}
// calculating the median.
// If number of elements
// is odd there is
// one middle element.
if (($n + $m) % 2 == 1)
return $median;
// Elements from a[] in the
// second half is an empty set.
if ($i == $n)
return (($median +
$b[$j]) / 2.0);
// Elements from b[] in the
// second half is an empty set.
if ($j == $m)
return (($median +
$a[$i]) / 2.0);
return (($median +
minimum($a[$i],
$b[$j])) / 2.0);
}
// Driver code
$a = array(900);
$b = array(10, 13, 14);
$n = count($a);
$m = count($b);
// we need to define the
// smaller array as the
// first parameter to make
// sure that the time complexity
// will be O(log(min(n,m)))
if ($n < $m)
echo ("The median is : " .
findMedianSortedArrays($a, $n,
$b, $m));
else
echo ("The median is : " .
findMedianSortedArrays($b, $m,
$a, $n));
// This code is contributed
// by Manish Shaw(manishshaw1)
?>
输出：
The median is : 13.5
另一种方法：相同的程序，但是返回合并数组中存在的中位数（(n + m) / 2 - 1位置）：
C++
// CPP code for finding median of the given two
// sorted arrays that exists in the merged array ((n+m) / 2 - 1 position)
#include<bits/stdc++.h>
using std::cout;
int maximum(int a, int b);
// Function to find median of given two sorted arrays
int findMedianSortedArrays(int *a, int n,
int *b, int m)
{
int min_index = 0, max_index = n, i, j;
while (min_index <= max_index)
{
i = (min_index + max_index) / 2;
j = ((n + m + 1) / 2) - i;
// if i = n, it means that Elements from a[] in
// the second half is an empty set. If j = 0, it
// means that Elements from b[] in the first half
// is an empty set. so it is necessary to check that,
// because we compare elements from these two groups.
// searching on right
if (i < n && j > 0 && b[j - 1] > a[i])
min_index = i + 1;
// if i = 0, it means that Elements from a[] in the
// first half is an empty set and if j = m, it means
// that Elements from b[] in the second half is an
// empty set. so it is necessary to check that,
// because we compare elements from these two groups.
// searching on left
else if (i > 0 && j < m && b[j] < a[i - 1])
max_index = i - 1;
// we have found the desired halves.
else
{
// this condition happens when we don't have
// any elements in the first half from a[] so
// we returning the last element in b[] from
// the first half.
if (i == 0)
return b[j - 1];
// and this condition happens when we don't have any
// elements in the first half from b[] so we
// returning the last element in a[] from the first half.
if (j == 0)
return a[i - 1];
else
return maximum(a[i - 1], b[j - 1]);
}
}
cout << "ERROR!!! " << "returning\n";
return 0;
}
// Function to find maximum
int maximum(int a, int b)
{
return a > b ? a : b;
}
// Driver code
int main()
{
int a[] = {900};
int b[] = { 10,13,14};
int n = sizeof(a) / sizeof(int);
int m = sizeof(b) / sizeof(int);
// we need to define the smaller array as the first
// parameter to make sure that the time complexity
// will be O(log(min(n,m)))
if (n < m)
cout << "The median is: "
<< findMedianSortedArrays(a, n, b, m);
else
cout << "The median is: "
<< findMedianSortedArrays(b, m, a, n);
return 0;
}
C
// C# code for finding median
// of the given two sorted
// arrays that exists in the
// merged array ((n+m) / 2 -
// 1 position)
using System;
class GFG
{
// Function to find maximum
static int maximum(int a,
int b)
{
return a > b ? a : b;
}
// Function to find median
// of given two sorted arrays
static int findMedianSortedArrays(ref int []a, int n,
ref int []b, int m)
{
int min_index = 0,
max_index = n, i, j;
while (min_index <= max_index)
{
i = (min_index + max_index) / 2;
j = ((n + m + 1) / 2) - i;
// if i = n, it means that
// Elements from a[] in the
// second half is an empty
// set. If j = 0, it means
// that Elements from b[]
// in the first half is an
// empty set. so it is
// necessary to check that,
// because we compare elements
// from these two groups.
// searching on right
if (i < n && j > 0 &&
b[j - 1] > a[i])
min_index = i + 1;
// if i = 0, it means that
// Elements from a[] in the
// first half is an empty set
// and if j = m, it means that
// Elements from b[] in the
// second half is an empty set.
// so it is necessary to check
// that, because we compare
// elements from these two
// groups. searching on left
else if (i > 0 && j < m &&
b[j] < a[i - 1])
max_index = i - 1;
// we have found the
// desired halves.
else
{
// this condition happens
// when we don't have any
// elements in the first
// half from a[] so we
// returning the last
// element in b[] from
// the first half.
if (i == 0)
return b[j - 1];
// and this condition happens
// when we don't have any
// elements in the first half
// from b[] so we returning the
// last element in a[] from the
// first half.
if (j == 0)
return a[i - 1];
else
return maximum(a[i - 1],
b[j - 1]);
}
}
Console.Write ("ERROR!!! " +
"returning\n");
return 0;
}
// Driver code
static void Main()
{
int []a = new int[]{900};
int []b = new int[]{10, 13, 14};
int n = a.Length;
int m = b.Length;
// we need to define the smaller
// array as the first parameter
// to make sure that the time
// complexity will be O(log(min(n,m)))
if (n < m)
Console.Write ("The median is: " +
findMedianSortedArrays(ref a, n,
ref b, m));
else
Console.Write ("The median is: " +
findMedianSortedArrays(ref b, m,
ref a, n));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
Python 3
# Python 3 code for finding median
# of the given two sorted arrays that
# exists in the merged array
# ((n+m) / 2 - 1 position)
# Function to find median of given
# two sorted arrays
def findMedianSortedArrays(a, n, b, m):
min_index = 0
max_index = n
while (min_index <= max_index):
i = (min_index + max_index) // 2
j = ((n + m + 1) // 2) - i
# if i = n, it means that Elements
# from a[] in the second half is
# an empty set. If j = 0, it means
# that Elements from b[] in the first
# half is an empty set. so it is necessary
# to check that, because we compare elements
# from these two groups. searching on right
if (i < n and j > 0 and b[j - 1] > a[i]):
min_index = i + 1
# if i = 0, it means that Elements from
# a[] in the first half is an empty set
# and if j = m, it means that Elements
# from b[] in the second half is an
# a[] in the first half is an empty set
# that, because we compare elements from
# these two groups. searching on left
elif (i > 0 and j < m and b[j] < a[i - 1]):
max_index = i - 1
# we have found the desired halves.
else:
# this condition happens when we don't have
# any elements in the first half from a[] so
# we returning the last element in b[] from
# the first half.
if (i == 0):
return b[j - 1]
# and this condition happens when we don't
# have any elements in the first half from
# b[] so we returning the last element in
# a[] from the first half.
if (j == 0):
return a[i - 1]
else:
return maximum(a[i - 1], b[j - 1])
print("ERROR!!! " , "returning\n")
return 0
# Function to find maximum
def maximum(a, b) :
return a if a > b else b
# Driver code
if __name__ == "__main__":
a = [900]
b = [10, 13, 14]
n = len(a)
m = len(b)
# we need to define the smaller array
# as the first parameter to make sure
# that the time complexity will be
# O(log(min(n,m)))
if (n < m):
print( "The median is: ",
findMedianSortedArrays(a, n, b, m))
else:
print( "The median is: ",
findMedianSortedArrays(b, m, a, n))
# This code is contributed
# by ChitraNayal
输出：
The median is: 13
时间复杂度：O(log(min(n(m, m))))，其中n和m是给定排序数组的大小。
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