> For the complete documentation index, see [llms.txt](https://garychang.gitbook.io/data-structure/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://garychang.gitbook.io/data-structure/3-search/sophisticated-sorting/3.3.1-quick-sort.md).

# 3.3.1 - Quick Sort

### 1. 演算法

採用Divide and Conquer策略。選定一個pivot key使得:

1. A\[1]\~A\[q-1] ≤ p.k
2.  A\[q+1]\~A\[n] ≥ p.k

此時資料被分成左右兩個sublist，再各自排列，當左右sublist排完，整個資料也就排序完成。

排序步驟範例:

$$A\[8]={5,9,8,2,3,6,4,7} \Rightarrow p.k=5$$ 

$$i$$ 找到大於p.k的位置， $$j$$找到小於p.k的位置 $$\Rightarrow A\[1]=9,\ A\[6]=4$$ 

$$swap \Rightarrow {5,\underline{4},8,2,3,6,\underline{9},7}$$ 

$$i$$ 找到大於p.k的位置， $$j$$找到小於p.k的位置 $$\Rightarrow A\[2]=8,\ A\[4]=3$$ 

$$swap \Rightarrow {5,4,\underline{3},2,\underline{8},6,9,7}$$ 

當 $$i≥j$$時，p.k和 $$A\[j]$$交換位置，此時 $$j=3$$ 

$$\Rightarrow {\[2,4,3],5,\[8,6,9,7]}$$ 

採用同樣方法`sort({2,3,4})`、`sort({8,6,9,7})`

```
void Sort(char *arr, int l, int u){
    if (l >= u) return; //判斷sorting結束
    int pk = arr[l];    //選pk
    int i = l;
    int j = u ;
    int temp;
    while (i<j){        //直到i>j
        while (arr[i]<pk) i++;
        while (arr[j]>pk) j--;
        if (i<j){     //swap
            temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }
    temp = arr[l];    //j與pk交換位置
    arr[l] = arr[j];
    arr[l] = temp;
    Sort(arr, l, j-1);  //左邊sublist
    Sort(arr, j+1, u);  //右邊sublist
}
```

### 2. 性質

* Time Complexity:

  * Best Case: $$O(nlogn)$$&#x20;

  &#x20;  $$\Rightarrow$$ p.k的位置恰將資料分割成兩等份

  &#x20;  $$T(n)=C\times n+T(\frac{n}{2})+T(\frac{n}{2})$$ C\*n為分割所花時間

  &#x20;  $$T(n)=2T(\frac{n}{2})+C\times n$$&#x20;

  &#x20;  $$T(\frac{n}{2})=2T(\frac{n}{4})+C\times\frac{n}{2}$$&#x20;

  &#x20;

  &#x20;  $$T(n)=2\[2T(\frac{n}{4})+C\times\frac{n}{2}]+C\times n=4T(\frac{n}{4})+2(C\times n)$$&#x20;

  &#x20;  $$T(n)=nT(1)+\log\_2n(C\times n)$$&#x20;

  * Worst Case: $$O(n^2)$$&#x20;

  &#x20;  $$\Rightarrow$$ p.k恰好是最大值或是最小值，使得分割無效果

  &#x20;  $$T(n)=C\times n+T(n-1)$$ C\*n為分割所花時間

  &#x20;  $$T(n)=C(n+(n-1))+T(n-2)$$&#x20;

  &#x20;  $$T(n)=C(n+...+2)+T(1)=\frac{C(n+2)(n-1)}{2}$$&#x20;

* Space Complexity: $$O(logn)$$\~ $$O(n)$$&#x20;

  &#x20;  $$\Rightarrow$$ 額外空間需求來自於recursion所需的stack，stack size取決於recursion call的次數。

  * Best Case: $$O(logn)$$&#x20;

  &#x20;  $$\Rightarrow$$ 每次p.k的位置恰將資料分割成兩等份，經過k次呼叫後只剩一筆Data， $$\frac{n}{2^k}=1,\ k=log\_2n$$&#x20;

  * Worst Case: $$O(n)$$&#x20;

  &#x20;  $$\Rightarrow$$ p.k恰好是最大值或是最小值，使得分割無效果，資料大小為n, n-1, n-2,...，共需n次呼叫

Quick sorting is a **unstable** sorting method.

$${3\_{p.k},\ 5,\ 5^*,\ 1,\ 2 } \Rightarrow \[1,\ 2],\ 3,\ \[5^*,\ 5]$$&#x20;


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