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排序算法的实现(归并,快排,堆排,希尔排序 O(N*log(N)))

      今天跟着左老师的视频,理解了四种复杂度为 O(N*log(N))的排序算法,以前也理解过过程,今天根据实际的代码,感觉基本的算法还是很简单的,只是自己写的时候可能一些边界条件,循环控制条件把握不好。

//对于一个int数组,请编写一个选择冒泡算法,对数组元素排序。
//给定一个int数组A及数组的大小n,请返回排序后的数组。
//测试样例:
//[1, 2, 3, 5, 2, 3], 6
//[1, 2, 2, 3, 3, 5]


#include <iostream> 
using namespace std;
#include<string>

void printResult(string str,int* A,int n)
{
    cout << str << "的结果:\n";
    for (int i = 0; i < n; i++)
    {
        cout << A[i] <<" ";
    }
    cout << endl;
}
void swap(int *a, int *b)
{
    int temp=*a;
    *a = *b;
    *b = temp;
}

//冒泡排序 O(n^2) 
class BubbleSort {
public:
    int* bubbleSort(int* A, int n) {
        // write code here
        for (int i = 0; i<n; i++)
        {
            for (int j = 0; j<n - i - 1; j++)
            {
                if (A[j]>A[j + 1])
                {
                    int temp = A[j];
                    A[j] = A[j + 1];
                    A[j + 1] = temp;
                }
            }
        }
        return A;
    }
};

//请编写一个选择排序算法 O(n^2) 
class SelectionSort {
public:
    int* selectionSort(int* A, int n) {
        // write code here
        int k = 0;
        for (int i = 0; i < n-1; i++)
        {
            k = i;
            for (int j = i; j < n; j++)
            {
                if (A[k]>A[j])
                {
                    k = j;
                }
            }
            if (k!=i)
            {
                int temp = A[i];
                A[i] = A[k];
                A[k] = temp;
            }
        }
        return A;
    }
};

//请编写一个插入算法 O(n^2) 
class InsertionSort
{
public:
    int* insertionSort(int* A, int n)
    {
        for (int i = 1; i < n; i++)
        {
            int temp = A[i];
            int j = i - 1;
            for (; j >= 0;j--)   //j前面的已经排好序,从后面往前比较,当没有比当前值大的时候bereak;
            {
                if (A[j]>temp)
                {
                    A[j + 1] = A[j];
                }
                else
                {
                    break;
                }
            }
            A[j + 1] = temp;
        }
        return A;
    }
};

//归并排序 O(N*log(N))
class MergeSort {
public:
    int* mergeSort(int* A, int n) {
        // write code here
        mergeSort(A, 0, n - 1);
        return A;
    }
    void mergeSort(int* A, int beg, int end)
    {
        if (beg < end)
        {
            int mid = beg + (end - beg) / 2;
            mergeSort(A, beg, mid);
            mergeSort(A, mid + 1, end);
            merge(A,beg,mid,end);
        }
        return;
    }
    void merge(int* A, int beg_, int mid_, int end_)
    {
        int *B = new int[end_ - beg_ + 1];
        int index1 = beg_;
        int index2 = mid_ + 1;
        int i = 0;
        while (index1<=mid_&&index2<=end_)
        {
            if (A[index1]<=A[index2])
            {
                B[i++] = A[index1++];
            }
            else
            {
                B[i++] = A[index2++];
            }
        }
        while (index1 <= mid_)
        {
            B[i++] = A[index1++];
        }
        while (index2<=end_)
        {
            B[i++] = A[index2++];
        }
        //memcpy(A,B,end_-beg_+1);
        for (int i = 0; i < end_ - beg_ + 1;i++)
        {
            A[beg_+i] = B[i];   //A[beg_++] 不能写,改变了输入参数
        }
        delete[] B;
    }
};

//快速排序 O(N*log(N))
#include <math.h>
class QuickSort {
public:
    int* quickSort(int* A, int n) {
        // write code here
        quickSort(A, 0, n - 1);
        return A;
    }
    void quickSort(int* A, int low, int high)
    {
        if (low <= high)
        {
            int part = partition(A, low, high);
            quickSort(A, low, part - 1);
            quickSort(A, part + 1, high);
        }
        return;
    }

    int partition(int* A, int low, int high)
    {
        int privotKey = A[low];    //基准元素
        while (low < high)
        {        //从表的两端交替地向中间扫描
            while (low < high  && A[high] >= privotKey) 
                --high;  //从high 所指位置向前搜索,至多到low+1 位置。将比基准元素小的交换到低端
            swap(&A[low], &A[high]);
            while (low < high  && A[low] <= privotKey) 
                ++low;
            swap(&A[low], &A[high]);
        }
        return low;
    }
};
class QuickSort2 {
public:
    int* quickSort(int* A, int n) {
        // write code here
        quickSort(A, 0, n - 1);
        return A;
    }
    void quickSort(int* A, int low, int high)
    {
        if (low <= high)
        {
            int randn = low + rand() % (high - low + 1);  //随机选择关键字的下标
            swap(&A[randn], &A[high]);                      //void swap(int* A,int index1,int index2) //最好都操作下标

            int part = partition(A, low, high);
            quickSort(A, low, part - 1);
            quickSort(A, part + 1, high);
        }
        return;
    }

    int partition(int* A, int low, int high) //O(N)
    {
        //int pivot = A[low];//很多随机选择放在这里面,而且是以值的形式确定,而非下标标记为关键字
        
        int pivot = low-1; //关键字的位置
        for (int i = low ; i <= high; i++)
        {
            if (A[i] <= A[high])
            {
                swap(&A[i], &A[++pivot]);  //感觉这样会把A数组前面的值覆盖?-->其实没有交换的效果就是把前面的交换到后面
            }
        }
        return pivot;
    }
};

//推排序  O(N*log(N))
class HeapSort {
public:
    int* heapSort(int* A, int n) {
        // write code here
        buildHeap(A, n); //初始时构建堆
        //从最后一个元素开始对序列进行调整
        for (int i = n - 1; i >= 0;i--)
        {
            swap(&A[0], &A[i]);
            heapAdjust(A,0,i);
        }
        return A;
    }

    void buildHeap(int* A, int size_A)
    {
        for (int i = (size_A)/ 2-1; i >= 0; i--)
        {
            heapAdjust(A,i,size_A);
        }
    }

    void heapAdjust(int* A, int root, int size_A) //大顶堆
    {
        int leftchild = 2 * root + 1;
        if (leftchild<size_A) //递归形式
        {
            int rightchild = leftchild + 1;
            if (rightchild<size_A)
            {
                if (A[leftchild]<A[rightchild])
                {
                    leftchild = rightchild;
                }
            }
            //leftchild为左右子节点中较大的结点
            if (A[root]<A[leftchild])
            {
                int temp = A[root];
                A[root] = A[leftchild];   //将较大结点值上移到根节点
                A[leftchild] = temp; //完成交换,子节点变为以前的根节点
                heapAdjust(A, leftchild, size_A);
            }
        }
        return;
    }
};
class HeapSort2 {
public:
    int* heapSort(int* A, int n) {
        // write code here
        buildHeap(A, n); //初始时构建堆
        //从最后一个元素开始对序列进行调整
        for (int i = n - 1; i >= 0; i--)
        {
            swap(&A[0], &A[i]);
            heapAdjust(A, 0, i);
        }
        return A;
    }

    void buildHeap(int* A, int size_A)
    {
        for (int i = (size_A - 1) / 2; i >= 0; i--)
        {
            heapAdjust(A, i, size_A);
        }
    }

    void heapAdjust(int* A, int root, int size_A)  //调整为大顶堆
    {
        int temp = A[root];
        int leftchild = 2 * root + 1;
        while (leftchild < size_A) //非递归形式
        {
            int rightchild = leftchild + 1;
            if (rightchild < size_A)
            {
                if (A[leftchild] < A[rightchild])
                {
                    leftchild = rightchild;
                }
            }
            //leftchild为左右子节点中较大的结点
            if (A[root] < A[leftchild])
            {
                A[root] = A[leftchild];   //将较大结点值上移到根节点
                root = leftchild;         //更新新的根节点
                leftchild = 2 * root + 1;        
            }
            else  //当前结点大于左右子节点则不需要调整
            {
                break;
            }        
            A[root] = temp; //完成交换,子节点变为以前的根节点
        }
        return;
    }
};

//希尔排序  O(N*log(N)) ---不稳定
class ShellSort {
public:
    int* shellSort(int* A, int n) {
        // write code here
        int dk = n / 2;
        while (dk>=1)
        {
            shellSort2(A,n,dk);
            dk /= 2;
        }
        return A;
    }
    void shellSort(int* A, int n, int dk)
    {
        for (int i = dk; i < n;i++)
        {
            int index = i; //当前访问的位置
            while (index>=dk)
            {
                if (A[index-dk]>A[index])
                {
                    swap(&A[index-dk],&A[index]); //交换不算最优,找到插入位置才交换
                    index -= dk;
                }
                else
                {
                    break;
                }
            }
        }
    }
    void shellSort2(int* A,int n,int dk)
    {
        for (int i = dk; i < n;i++)
        {
            if (A[i]<A[i-dk]) //找到插入位置
            {
                int x = A[i];//复制哨兵
                A[i] = A[i - dk];
                int j = i - dk; //从该位置向前查找
                while (x<A[j]&&j>=0) //防止j越界
                {
                    A[j] = A[j - dk];
                    j -= dk; //向前移动
                }
                A[j + dk] = x;// 插入到正确位置
            }
        }
    }

};


#define N 13
int main()
{
    //待排数据输入方式:
        /*int N = 0;
        cout << "排序数据个数:\n";
        cin >> N;
        int* A = new int[N];
        cout << "请输入待排序的数据:\n";
        for (int i = 0; i < N; i++)
        {
        cin >> A[i];
        }*/
    //数据直接给定    
        int B[N] = { 1, 6, 3, 5, 2, 4 };
        int C[13] = { 54, 35, 48, 36, 27, 12, 44, 44, 8, 14, 26, 17, 2 };
        int* A = C;

    //从文件中读取,大量数据,计算时间复杂度
        
    printResult("待排原始数据:", C, N);

    BubbleSort bubble;
    bubble.bubbleSort(A,N);
    printResult("bubbleSort", A, N);

    SelectionSort select;
    select.selectionSort(A, N);
    printResult("selectSort", A, N);

    InsertionSort insert;
    insert.insertionSort(A, N);
    printResult("InsetSort", A, N);

    MergeSort merge;
    merge.mergeSort(A, N);
    printResult("MergeSort", A, N);

    QuickSort qucik;
    qucik.quickSort(A, N);
    printResult("QucikSort",A,N);

    QuickSort2 qucik2;
    qucik2.quickSort(A, N);
    printResult("QucikSort2", A, N);

    HeapSort heap;
    heap.heapSort(A, N);
    printResult("heapSort", A, N);

    HeapSort2 heap2;
    heap2.heapSort(A, N);
    printResult("heapSort2", A, N);


    ShellSort shell;
    shell.shellSort(A,N);
    printResult("shellSort", A, N);

    return 0;
}

 

posted @ 2016-10-04 15:30  ranjiewen  阅读(702)  评论(0编辑  收藏  举报