要点:
1、大于基准pv的数放在一边,小于pv的放在另一边.
2、Partition函数返回基准数的位置pivot
对基准的左边快排Quick(array, begin, pivot-1);
对基准的右边快排Quick(array, pivot+1, end);
3、递归求解,直到不满足begin<end
#include <iostream>
using namespace std;
template <typename T>
static void Swap(T& a, T& b)
{
T c(a);
a = b;
b = c;
}
template <typename T>
static int Partition(T array[], int begin, int end, bool min2max)
{
T pv = array[begin];
while( begin < end )
{
while( (begin < end) && (min2max ? (array[end] > pv) : (array[end] < pv)) )
{
end--;
}
Swap(array[begin], array[end]);
while((begin < end) && (min2max ? (array[begin] <= pv) : (array[begin] >= pv)))
{
begin++;
}
Swap(array[begin], array[end]);
}
array[begin] = pv;
return begin;
}
template <typename T>
static void Quick(T array[], int begin, int end, bool min2max)
{
if(begin < end)
{
int pivot = Partition(array, begin, end, min2max);
Quick(array, begin, pivot-1, min2max);
Quick(array, pivot+1, end, min2max);
}
}
template <typename T>
static void Quick(T array[], int len, bool min2max = true)
{
Quick(array, 0, len-1, min2max);
}
int main()
{
int array[] = {9,7,2,1,4,3,6,5,8,9};
Quick(array, 10, false);
for(int i=0; i<10; i++)
{
cout << array[i] << endl;
}
cout << endl;
Quick(array, 10, true);
for(int i=0; i<10; i++)
{
cout << array[i] << endl;
}
return 0;
}