插入排序和分治排序

What’s more important than performance?

> modularity

> correctness

> maintainability

> functionality

> robustness

> user-friendliness

> programmer time

> simplicity

> extensibility

> reliability

Why study algorithms and performance?

> Algorithms help us to understand scalability.

> Performance often draws the line between what is feasible and what is impossible.

> Algorithmic mathematics provides a language for talking about program behavior.

> The lessons of program performance generalize to other computing resources. 

> Speed is fun!

插入排序法（少量資料排序較好，是一種增量排序方法）：O(n2)

說明：縮排代表程式結構，三角形代表注釋，箭頭表示賦值。

Running time

• The running time depends on the input: an already sorted sequence is easier to sort.

• Parameterize the running time by the size of the input, since short sequences are easier to sort than long ones.

• Generally, we seek upper bounds on the running time, because everybody likes a Guarantee.

Kinds of analyses

Worst-case: (usually)

• T(n) = maximum time of algorithm on any input of size n.

Average-case: (sometimes)

• T(n) = expected time of algorithm over all inputs of size n.

• Need assumption of statistical distribution of inputs.

Best-case: (bogus)

• Cheat with a slow algorithm that works fast on some input

 

分治排序：O(nlogn)（是一種分結合并演算法或遞迴演算法）

演算法：

時間複雜度：

可以證明，其複雜度為O(nlogn)。

下面看一個例子：

有這樣一組資料，{5,4,1,22,12,32,45,21}，如果對它進行合并排序的話，首先將它從中間分開，這樣，它就被分成了兩個數組{5,4,1,22} {12,32,45,21}.

對這兩個數組，也分別進行這樣的操作，逐步的劃分，直到不能再劃分為止(每個子數組只剩下一個元素)，這樣，劃分的過程就結束了。

劃分的過程如所示：

　　接下來，我們進行合併作業，依照，劃分過程是從上到下進行的，而合并的過程是從下往上進行的，例如中,最下層{5},{4}這兩個數組，如果按升序排列，將他們合并後的數組就是{4,5}。{1}，{22}這兩個子數組合并後是{1,22}。而{4,5}與{1,22}，這兩個數組同屬一個分支，他們也需要進行合并，由於這兩個子數組本身就是有序的，所以合并的過程就是，每次從待合并的兩個子數組中選取一個最小的元素，然後把這個元素放到合并後的數組中，前面兩個數組合并後就是{1,4,5,22}。依次類推，直到合并到最上層結束，這是資料的排序已經完成了。

合并的過程如所示。這個過程是從下往上的。

C語言實現代碼如下：

 1#include <stdlib.h> 2 3//合并過程 4void merge(int data[],int start,int mid,int end){ 5 6 7 int *tmpLeft,*tmpRight; 8 int leftSize,rightSize; 9 int l,r,j;1011    printArray(data,8);12    printf("\n");13    l = 0;14    r = 0;15    j = 0;16    leftSize = mid - start + 1;17    rightSize = end - mid;1819    tmpLeft = (int *)malloc(leftSize * sizeof(int));20    tmpRight = (int *)malloc(rightSize * sizeof(int));2122 while(j < leftSize){23        tmpLeft[j] = data[start + j];24        j++;25    }2627    j = 0;2829 while(j < rightSize){30        tmpRight[j] = data[mid + 1 + j];31        j++;32    }3334    j = 0;3536 while(l < leftSize && r < rightSize){37 if(tmpLeft[l] < tmpRight[r]){3839            data[start + j++] = tmpLeft[l++];4041        }else{4243            data[start + j++] = tmpRight[r++];44        }45    }4647 while(l < leftSize){48        data[start + j++] = tmpLeft[l++];49    }5051 while(r < rightSize){52        data[start + j++] = tmpRight[r++];53    }5455    free(tmpLeft);56    free(tmpRight);57}585960void merge_sort(int data[],int start,int end){6162 int mid;63 if(start < end){64 //將數組劃分65        mid = (start + end) / 2;66        merge_sort(data,start,mid);67        merge_sort(data,mid + 1,end);68 //合并劃分後的兩個數組69        merge(data,start,mid,end);70    }7172}