两列有序数 求第K个

时间复杂度 O(log(m + n))

#include "stdio.h"                                   
char arr1[4]={1,2,3,4};                              
char arr2[5]={5,6,7,8,9};                            
                                                     
#define MIN(x,y) (x > y) ? y : x                     
char findk(int i,int j,int k)                        
{                                                    
        char midx=0;                                 
        char midy=0;                                 
                                                     
        if(i >= sizeof(arr1)) return arr2[j+k-1];    
                                                     
        if(j >= sizeof(arr2)) return arr1[i+k-1];    
                                                     
        if(k==1) return MIN(arr1[i],arr2[j]);        
                                                     
        if(i + k/2 -1 < sizeof(arr1))                
        {                                            
            midx = arr1[i+k/2 -1];                   
        }                                            
        else                                         
        {                                            
            midx=0x4f;                               
        }                                            
                                                     
        if(j+ k/2 -1 < sizeof(arr2))                 
        {                                            
            midy = arr2[j+k/2 -1];                   
        }                                            
        else                                         
        {                                            
            midy = 0x4f;                             
        }                                            
        if(midx < midy)                              
        {                                            
            return findk(i+k/2,j, k-k/2);            
        }                                            
        else                                         
        {                                            
            return findk(i,j+k/2, k-k/2);            
        }                                            
}