编程之美（数组求最大最小值的集中方法）

#include <iostream>
using namespace std;
#include <assert.h>
//分治思想求数组最大最小值
void MaxandMin(int* a,int start,int end,int &Max,int &Min)
{
if(end-start<=1)
{
if(a[start] > a[end])
{
Max = a[start];
Min = a[end];
}
else
{
Max = a[end];
Min = a[start];
}
return;
}
int RightMax=0;
int RightMin=0;
int LeftMax=0;
int LeftMin=0;
MaxandMin(a,start,(start+end)/2,LeftMax,LeftMin);
MaxandMin(a,(start+end)/2+1,end,RightMax,RightMin);

Max = (RightMax > LeftMax) ? RightMax : LeftMax;
Min = (LeftMin < RightMin) ? LeftMin : RightMin;
}
//暴力方法
void MaxandMin2(int *a,int len,int &max,int &min)
{
//if(a == NULL || len <= 0)
//  cout<<"error!"<<endl;
assert(len > 0); //if(真) 程序运行；else 报错&&终止程序
max=a[0];
min=a[0];
for(int i(1);i<=len-1;++i)
{
if(max < a[i])
max = a[i];
if(min > a[i])
min = a[i];
}

}
// 遍历一遍数组使得下标为偶数的元素较下标为奇数的元素大,再分别求出最大值和最小值
//比较次数为3*N/2次
void MaxandMin3(int *a,int len,int &max,int &min)
{
if(a == NULL || len<= 0 )
return;
for(int i(0);i<len;i+=2)
{
if(i<len && a[i] < a[i+1])
{
int temp = a[i+1];
a[i+1] = a[i];
a[i] = temp;
}
}
max=a[0];
for(int j(2);j<len;j+=2)
{
if(max < a[j])
{
max = a[j];
}
}
min = a[1];
for(int k(3);k<len;k+=2)
{
if(min > a[k])
{
min = a[k];
}
}
}

//
void MaxandMin4(int *a,int len,int &max,int &min)
{
if(len > 0 && a != NULL)
{
if(len == 1)
{
max = min = a[0];
}
if(len == 2)
{
if(a[0] > a[1])
{
max = a[0];
min = a[1];
}
}

//
if(a[0] > a[1])
{
max = a[0];
min = a[1];
}
else
{
max = a[1];
min = a[0];
}
for(int i(2);i<len;i+=2)
{
if(i+1 < len && a[i] > a[i+1])
{
if(max < a[i])
{
max = a[i];
}
if(min > a[i+1])
{
min = a[i+1];
}
}
else if(i+1 < len && a[i] < a[i+1])
{
if(max < a[i+1])
{
max = a[i+1];
}
if(min > a[i])
{
min = a[i];
}
}
else
{
if(max < a[i])
{
max = a[i];
}
if(min > a[i])
{
min = a[i];
}
}
}
}
}

int main()
{
int a[10]={1,2,3,4,9,8,7,6,5,0};
int max=0;
int min =0;
MaxandMin(a,0,9,max,min);
cout<<"max="<<max<<endl<<"min="<<min<<endl;
MaxandMin2(a,10,max,min);
cout<<"max="<<max<<endl<<"min="<<min<<endl;
MaxandMin3(a,10,max,min);
cout<<"max="<<max<<endl<<"min="<<min<<endl;
MaxandMin4(a,10,max,min);
cout<<"max="<<max<<endl<<"min="<<min<<endl;
return 0;
}