如何用c++的分治法求数组最大最小值

http://www.cppblog.com/liyuxia713/archive/2009/04/14/79855.html
算法思想：先相邻两个两个比较，较大的放入数组max[],较小的放入数组min[],然后从max[]数组求出最大，min[]数组求出最小即可。 可以证明这是效率最高的算法，不能进一步改进。

#include
#define n 11
#define m ((n+1)/2)
using namespace std；

void main(void)
{
int num[] = {11,2,3,4,6,5,7,8,9,10,20};
//int n = sizeof(num)/sizeof(num[0]);
//int m = (n+1)/2;
int max[m] , min[m];
int k = 0, j = 0;

if(n/2 != 0) max[m-1] = min[m-1] = num[n-1];

for (int i=0; i < n-1; i = i+2)
{
if (num[i] >= num[i+1])
{
max[j++] = num[i];
min[k++] = num[i+1];
}
else
{
max[j++] = num[i+1];
min[k++] = num[i];
}

}

for( i=0; i< m; i++)
{
cout << "max[" << i << "] = " << max[i] << "\t";
cout << "min[" << i << "] = " << min[i] <}

int MAX = max[0];
int MIN = min[0];

for ( j = 1; j < m; j++)
{
if (max[j] > MAX) MAX = max[j];
if (min[j] < MIN) MIN = min[j];
}

cout << "MAX = " << MAX << ", MIN = " << MIN <}

我们可以把一个数组分为两块，再分别求这两块的最大值（或最小值），然后再比较这两个值，最大（小）者即为所求。

鉴于楼主一分不给，就写个伪代码了。

merge_large(a : array, length: integer) :integer
begin
large1, large2 :integer;
large1 = merge_large(a, length/2);
large2 = merge_large(a+length/2, length);
return large1 > large2 ? large1 : large2;
end