排序算法測試-歸併排序

1、測試代碼

#include <iostream>

using namespace std;
int g_loop = 0;    /* 循環計數 */
int g_move = 0;    /* 數據移動次數 */

void output_info(int *buff,int len, int flag)
{
    int i;

    if(0 == flag)
    {
        cout << "buff: ";
    }
    else
    {
        cout << "temp: ";
    }

    for(i = 0; i < len; ++i)
    {
        cout << *(buff + i) << " ";
    }

    cout << endl;
}

void merge(int arr[], int left,int mid,int right,int temp[])
{
    int i = left;     // 左序列指針
    int j = mid + 1;  // 右序列指針
    int pos = 0;      // 臨時數組指針

    while (i <= mid && j <= right)
    {
        if(arr[i] <= arr[j])
        {
            temp[pos++] = arr[i++];
        }
        else
        {
            temp[pos++] = arr[j++];
        }
    }
    while(i <= mid)    // 將左數組剩餘元素放進temp中
    {
        temp[pos++] = arr[i++];
    }
    while(j <= right)  // 將右數組剩餘元素放進temp中
    {
        temp[pos++] = arr[j++];
    }

    //將temp中的元素全部拷貝到原數組中
    output_info(arr, 10, 0);

    pos = 0;
    while(left <= right)
    {
        arr[left++] = temp[pos++];
        ++g_move;
    }

    output_info(temp, 10, 1);
}

void spread_out(int *buff, int left,int right, int *temp)
{
    int mid;

    ++g_loop;

    if(right > left)
    {
        mid = left + (right - left) / 2;

        spread_out(buff, left, mid, temp);
        spread_out(buff, mid + 1, right, temp);

        merge(buff, left, mid, right, temp);
    }

    if(right - left == 1)       /* 左右數組中只有一個數據 */
    {
        cout<< "left=" << *(buff + left) << " right=" << *(buff + right) << " end"<<endl;
    }
    else if(right - left == 0)  /* 左數組中有一個數據，右數組空 */
    {
        cout<< "left=" << *(buff + left) << " end"<<endl;
    }

    cout<<endl;
}

/* 歸併排序 */
int main()
{
    int array[10]= {10,9,8,7,6,5,4,3,2,1};
    int temp[10]= {0};

    spread_out(array, 0 ,9, temp);

    output_info(array, 10, 0);

    cout << endl;
    cout << "move=" << g_move << endl;
    cout << "loop=" << g_loop << endl;

    return 0;
}

2、測試log

left=10 end

left=9 end

buff: 10 9 8 7 6 5 4 3 2 1
temp: 9 10 0 0 0 0 0 0 0 0
left=9 right=10 end

left=8 end

buff: 9 10 8 7 6 5 4 3 2 1
temp: 8 9 10 0 0 0 0 0 0 0

left=7 end

left=6 end

buff: 8 9 10 7 6 5 4 3 2 1
temp: 6 7 10 0 0 0 0 0 0 0
left=6 right=7 end

buff: 8 9 10 6 7 5 4 3 2 1
temp: 6 7 8 9 10 0 0 0 0 0

left=5 end

left=4 end

buff: 6 7 8 9 10 5 4 3 2 1
temp: 4 5 8 9 10 0 0 0 0 0
left=4 right=5 end

left=3 end

buff: 6 7 8 9 10 4 5 3 2 1
temp: 3 4 5 9 10 0 0 0 0 0

left=2 end

left=1 end

buff: 6 7 8 9 10 3 4 5 2 1
temp: 1 2 5 9 10 0 0 0 0 0
left=1 right=2 end

buff: 6 7 8 9 10 3 4 5 1 2
temp: 1 2 3 4 5 0 0 0 0 0

buff: 6 7 8 9 10 1 2 3 4 5
temp: 1 2 3 4 5 6 7 8 9 10

buff: 1 2 3 4 5 6 7 8 9 10

move=34
loop=19

3、算法分析 

• 非原地排序算法；
• 穩定排序算法；
• 空間複雜度 O（n）；
• 時間複雜度 O（nlogn）。