有序数组中找中位数

原文：Median of two sorted arrays

题目：两个有序数组A和B，大小都是n，寻找这两个数组合并后的中位数。时间复杂度为O(logn)。
中位数：如果数组的个数是奇数，那么中位数的值就是有序时处于中间的数；如果数组个数是偶数的，那么就是有序时中间两个数的平均值。

方法一：合并时计数

使用Merge Sort时的Merge操作，比较两个数组时候计数，当计数达到n时，就可以得到中位数，在归并的数组中，中位数为下标n-1和n的两个数的平均值。
时间复杂度O(n)。

#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;          
        }          
        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");    
  
    return 0;  
}  

方法二：比较两个数组的中位数

ar1[]和ar2[]为输入的数组

算法过程：

1.得到数组ar1和ar2的中位数m1和m2

2.如果m1==m2，则完成，返回m1或者m2

3.如果m1>m2，则中位数在下面两个子数组中

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.如果m1<m2，则中位数在下面两个子数组中

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.重复上面的过程，直到两个子数组的大小都变成2

6.如果两个子数组的大小都变成2，使用下面的式子得到中位数：Median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2

时间复杂度：O(logn)。

#include <stdio.h>      
  
/* Utility functions */  
int max(int x, int y)   
{       
    return x > y? x : y;  
}    
  
int min(int x, int y)  
{     
    return x > y? y : x;   
}  
  
/* 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];   
}    
  
/*  
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 m1; /* For median of ar1 */    
    int m2; /* For median of ar2 */      
  
    /* 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;    
  
    m1 = median(ar1, n); /* get the median of the first array */    
    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);        
        else        
            return getMedian(ar1 + n/2, ar2, n - n/2);     
    }        
  
    /* if m1 > m2 then median must exist in ar1[....m1] and ar2[m2...] */      
    else     
    {      
        if (n % 2 == 0)          
            return getMedian(ar2 + n/2 - 1, ar1, n - n/2 + 1);          
        else            
            return getMedian(ar2 + n/2, ar1, n - 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;   
}   

方法三：通过二分查找法来找中位数

基本思想是：假设ar1[i]是合并后的中位数，那么ar1[i]大于ar1[]中前i-1个数，且大于ar2[]中前j=n-i-1个数。通过ar1[i]和ar2[j]、ar2[j+1]两个数的比较，在ar1[i]的左边或者ar1[i]右边继续进行二分查找。对于两个数组 ar1[] 和ar2[], 先在 ar1[] 中做二分查找。如果在ar1[]中没找到中位数, 继续在ar2[]中查找。

算法流程：
1) 得到数组ar1[]最中间的数，假设下标为i.
2) 计算对应在数组ar2[]的下标j，j = n-i-1
3) 如果 ar1[i] >= ar2[j] and ar1[i] <= ar2[j+1]，那么 ar1[i] 和 ar2[j] 就是两个中间元素，返回ar2[j] 和 ar1[i] 的平均值
4) 如果 ar1[i] 大于 ar2[j] 和 ar2[j+1] 那么在ar1[i]的左部分做二分查找(i.e., arr[left ... i-1])
5) 如果 ar1[i] 小于 ar2[j] 和 ar2[j+1] 那么在ar1[i]的右部分做二分查找(i.e., arr[i+1....right])
6) 如果到达数组ar1[]的边界(left or right)，则在数组ar2[]中做二分查找

时间复杂度：O(logn)。

#include <stdio.h>      
  
/* A recursive function to get the median of ar1[] and ar2[] using binary search */  
int getMedianRec(int ar1[], int ar2[], int left, int right, int n)   
{      
    int i, j;   
  
    /* We have reached at the end (left or right) of ar1[] */    
    if(left > right)        
        return getMedianRec(ar2, ar1, 0, n-1, n);    
  
    i = (left + right)/2;      
    j = n - i - 1;  /* Index of ar2[] */      
  
    /* Recursion terminates here.*/     
    if (ar1[i] > ar2[j] && (j == n-1 || ar1[i] <= ar2[j+1]))     
    {        
        /*ar1[i] is decided as median 2, now select the median 1         
        (element just before ar1[i] in merged array) to get the average of both*/     
        if (ar2[j] > ar1[i-1] || i == 0)          
            return (ar1[i] + ar2[j])/2;        
        else            
            return (ar1[i] + ar1[i-1])/2;    
    }  
  
    /*Search in left half of ar1[]*/     
    else if (ar1[i] > ar2[j] && j != n-1 && ar1[i] > ar2[j+1])       
        return getMedianRec(ar1, ar2, left, i-1, n);     
  
    /*Search in right half of ar1[]*/      
    else /* ar1[i] is smaller than both ar2[j] and ar2[j+1]*/    
        return getMedianRec(ar1, ar2, i+1, right, n);   
}  
  
/*  
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)   
{     
    // If all elements of array 1 are smaller then     
    // median is average of last element of ar1 and first element of ar2      
    if (ar1[n-1] < ar2[0])      
        return (ar1[n-1]+ar2[0])/2;       
  
    // If all elements of array 1 are smaller then       
    // median is average of first element of ar1 and       
    // last element of ar2      
    if (ar2[n-1] < ar1[0])      
        return (ar2[n-1]+ar1[0])/2;    
    
    return getMedianRec(ar1, ar2, 0, n-1, n);   
}  
  
/* 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");    
  
    return 0;   
}  

原文地址：http://www.geeksforgeeks.org/archives/2105