C语言描述算法

[vistalong]

用分治的方法设计算法，在数组A[1,N]中同时寻找最大值和最小值（假设n为2的方幂）用C语言描述算法并分析时间复杂度
我的思路是这样先把数组平分成两端分别求这两端的最大值和最小值在进行比较 求大家帮助实现

[WaterWalker]

minmax( A[1-N], int b, int e, int *min, int *max )
{
if ( e - b == 1 ){
*min = A[b] > A[e] ? A[e] : A[b]
*max = A[b] > A[e] ? A[b] : A[e]
return;
}

int min1, max1, min2, max2;
minmax(A, b, (b + e) / 2, &min1, &max1);
minmax(A, (b+e) / 2 + 1, e, &min2, &max2);

*min = minimal(min1, min2)
*max = maxest(max1, max2);

return;

}

T = O(n)

[qiannv]

才学,不会,路过.

[pptor]

#include <stdio.h>
int MAX(int a,int b)
{
return a>b?a:b;
}
int MIN(int a,int b)
{
return a<b?a:b;
}
//search max from a[] between low and high
int MaxSearch(int *a,int low,int high)
{
if(low==high) return a[low];
int middle=(low+high)/2;
return MAX(MaxSearch(a,low,middle),MaxSearch(a,middle+1,high));
}
//search min from a[] between low and high
int MinSearch(int *a,int low,int high)
{
if(low==high) return a[low];
int middle=(low+high)/2;
return MIN(MinSearch(a,low,middle),MinSearch(a,middle+1,high));
}
//when b is true search max from a[] between low and high,
//when b is false search min from a[] between low and high
int MAXMIN(int *a,int low,int high,bool b)//b true max ,false min
{
if(b) return MaxSearch(a,low,high);
else return MinSearch(a,low,high);
}
int main()
{
int a[20]={20,1984,1865,1795,1645,1594,14852,13455,12364,1135,10941,914,800,736,665,541,436,395,214,196};
int b;
b=MAXMIN(a,0,19,true);
printf("MAX=%d\n",b);
b=MAXMIN(a,0,19,false);
printf("MIN=%d\n",b);
return 1;
}

[vistalong]

非常感谢大家提供思路