0. 写作目的

好记性不如烂笔头。不同语言实现。（之前都是C++或者C语言实现的QuickSort，今天使用python3来实现）

1. 算法思想

对输入的数组 a[i , j] 进行排序:

1) 分解：将输入的数组分解成三部分—— a[i , k - 1], a[k] 和 a[k + 1 , j],使得a[i , k - 1]中的元素不大于(或者小于)a[k],a[k + 1, j]中的元素大于(或者不小于)a[k];

2) 递归求解：对于每部分递归调用快速排序算法，进行排序;

2. C++ 实现(模板函数)

template<class T>
int QuickSort_partition(T a[], int left, int right)
{
    T reference = a[left];
    while(left < right){
        while( (a[right] >= reference) && (left < right) ) right--;
        if( left < right ) a[left++] = a[right];
        while( (a[left] < reference) && (left < right) ) left++;
        if( left < right ) a[right--] = a[left];
    }
    a[right] = reference;
    return right;
}

template<class T>
void QuickSort(T a[], int left, int right)
{
    if(left < right){
        int middle_index = QuickSort_partition(a, left, right);
        QuickSort(a, left, middle_index - 1);
        QuickSort(a, middle_index + 1, right);
    }
}

3. python3实现

class QuickSortClass:
    def __init__(self, a:'list'):
        self.a = a

    def QuickSort_partition(self, left, right):
        reference = a[left]
        while (left < right):
            while ((a[right] >= reference) and (left < right)):
                right -= 1
            if (left < right):
                a[left] = a[right]
                left +=1
            while ((a[left] < reference) and (left < right)):
                left += 1
            if (left < right):
                a[right] = a[left]
                right -= 1
        a[right] = reference
        return right
    def QuickSort(self, left, right):
        if(left < right):
            middle_value_index = self.QuickSort_partition(left, right)
            self.QuickSort(left, middle_value_index - 1)
            self.QuickSort(middle_value_index + 1, right)

def test1():
    a = [3, 4, 13, 2, 13, 1]
    return a

if __name__ == "__main__":
    a = test1()
    print(a)
    quickSort = QuickSortClass(a)
    quickSort.QuickSort(0, len(a) - 1)
    print(quickSort.a)

[Reference]

[1] （博主的博客园，暂时不更新了）https://www.cnblogs.com/renchen/p/5935685.html