有两个大小分别为n的排序数组A和B。编写一个算法，求合并上述两个数组（即长度为2n的数组）后得到的数组的中值。复杂度应该是O（log（n））。
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注： 因为我们寻找的中位数集的大小是偶数（2n），所以我们需要取中间两个数字的平均值，并返回平均值的下限。 方法1（合并时简单计数） 使用合并排序的合并过程。在比较两个数组的元素时跟踪计数。如果计数变成n（对于2n个元素），我们就达到了中值。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 <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[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 <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[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" ); } } |
蟒蛇3
# 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" ); } } |
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 [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 $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. ?> |
Javascript
<script> // 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" ); </script> |
输出：
Median is 16
时间复杂性： O（n）
辅助空间： O（1） 方法2（通过比较两个数组的中间值） 该方法首先获取两个排序数组的中间值，然后比较它们。 让ar1和ar2作为输入数组。 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
例如：
ar1[] = {1, 12, 15, 26, 38} ar2[] = {2, 13, 17, 30, 45}
对于上述两个阵列m1=15和m2=17 对于上述ar1[]和ar2[]，m1小于m2。所以中位数出现在下面的两个子数组中。
[15, 26, 38] and [2, 13, 17]
让我们对上述两个子阵列重复该过程：
m1 = 26 m2 = 13.
m1大于m2。这样子阵列就变成了
[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
实施：
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[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<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[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 |
Javascript
<script> // A Javascript program to divide and conquer based // efficient solution to find // median of two sorted arrays // of same size. /* 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(a,b,startA,startB,endA,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 */ let m1 = median(a, startA, endA); /* get the median of the second array */ let 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 */ function median(arr,start,end) { let 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 let ar1 = [ 1, 2, 3, 6 ]; let ar2 = [ 4, 6, 8, 10 ]; let n1 = ar1.length; let n2 = ar2.length; if (n1 != n2) { document.write( "Doesn't work for arrays " + "of unequal size<br>" ); } else if (n1 == 0) { document.write( "Arrays are empty.<br>" ); } else if (n1 == 1) { document.write((ar1[0] + ar2[0]) / 2+ "<br>" ); } else { document.write( "Median is " + getMedian( ar1, ar2, 0, 0, ar1.length - 1, ar2.length - 1)+ "<br>" ); } // This code is contributed by avanitrachhadiya2155 </script> |
输出：
Median is 5
时间复杂性： O（logn）
辅助空间： O（1） 算法范式：分而治之
方法3（采用无额外空间的union）
这种方法的工作原理是在没有额外空间的情况下对两个数组进行并集，然后对它们进行排序。
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].
实施：
C++
// CPP program for the above approach #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 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 |
蟒蛇3
# 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 |
Javascript
<script> /* 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) { let j = 0; let i = n - 1; while (ar1[i] > ar2[j] && j < n && i > -1) { let temp = ar1[i]; ar1[i] = ar2[j]; ar2[j] = temp; i--; j++; } ar1.sort( function (a, b){ return a - b}); ar2.sort( function (a, b){ return a - b}); return parseInt((ar1[n - 1] + ar2[0]) / 2, 10); } let ar1 = [ 1, 12, 15, 26, 38 ]; let ar2 = [ 2, 13, 17, 30, 45 ]; let n1 = 5; let n2 = 5; if (n1 == n2) document.write( "Median is " + getMedian(ar1, ar2, n1)); else document.write( "Doesn't work for arrays of unequal size" ); </script> |
输出
Median is 16
时间复杂性： O（nlogn）
辅助空间： O（1）
两个大小不同的排序数组的中值 参考资料： 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.pdfds3etph5wn 如果您发现上述代码/算法不正确，请写下评论，或者寻找其他方法来解决相同的问题。
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