给定两个有序的n长度的数组，如何找出这两个数组合并后的中位数？

给定两个有序的n长度的数组，如何找出这两个数组合并后的中位数？

（1） o(n)的时间复杂度。这个就像两个有序链表合并一样，设定一个变量，记录合并数的索引，一旦等于n，即停止，当前值卫中位数。

（2）log（n）时间复杂度。通过比较两个数组中位数的大小，可以把查找范围缩小一半。

#include <iostream>
using namespace std;

//log(n) time;
int get_median2(int * a, int * b, int begin_a , int begin_b, int end_a ,int end_b){

	int a_median_index = 0 ;
	int b_median_index = 0;

	int result = 0;

	a_median_index = (begin_a + end_a)/2;
	b_median_index = (begin_b + end_b)/2;
    
	if (a[a_median_index] == b[b_median_index]){
		return a[a_median_index];
	}

	if (begin_a == end_a & begin_b == end_a){
		return a[begin_a] > b[begin_b] ? b[begin_b]:a[begin_a]; 
	}

	if (a[a_median_index] < b[b_median_index]){
		if ((begin_a + end_a +1) %2 == 0){
			result = get_median2(a,b, a_median_index + 1, begin_b, end_a, b_median_index);
		}

		else{
			result = get_median2(a,b, a_median_index, begin_b, end_a, b_median_index);
		}

		return result;
		
	}
	if (a[a_median_index] > b[b_median_index]){
		if ((begin_b + end_b +1) %2 == 0){
			result = get_median2(a,b, begin_a, b_median_index + 1, a_median_index, end_b);
		}

		else{
			result = get_median2(a,b, begin_a, b_median_index, a_median_index, end_b);
		}

		return result;
	}
	
	
}


// o(n) time;
int get_median(int * a, int * b, int m ,int n){

	int median_index = (m + n)/2;

	int i = 0, j =0 , k = 0;

	while(i < m & j< n){
		if (a[i] < b[j]){			
			k++;
			if (k == median_index)
				return a[i];
			i++;
		}
		else if ( a[i] == b[j]){			
			 k += 2;
			 if (k == median_index)
				 return a[i];
			 i++;
			 j++;
		}
		else{			
			k++;
			if (k == median_index)
				return b[j];
			j++;
		}		
	}

	
}


int main(){
	int a[5] = {1,3,5,7,9};
	int b[5] = {2,4,6,8,10};
	int result1 = get_median(a,b,5,5);
	int result2 = get_median2(a,b,0,0 ,4,4);
	cout <<"result1 : " << result1 << endl;
	cout <<"result2 : " << result2 << endl;
	system("pause()");
	return 0;
	
}