再谈快速排序

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本篇博客是在快速排序基础上优化的。

算法思想

快速排序是基于分治模式构思的，具体描述如下：

分解：数组A[p..r]被划分成两个（可能空）子数组A[p..q-1]和A[q+1..r]，使得A[p..q-1]中的每个元素都小于等于A(q)，而且，小于等于A[q+1..r]中的元素。下标q也在这个划分过程中进行计算。
解决：通过递归调用快速排序，对子数组A[p..q-1]和A[q+1..r]排序。
合并：因为两个子数组是就地排序的，将它们合并不需要操作：整个数组A[p..r]已排序。

// QUICKSORTQUICKSORT(A, p, r)   if p < r      then  q = PARTITION(A, p, r)            QUICKSORT(A, p, q-1)            QUICKSORT(A, q+1, r)// PARTITIONPARTITION(A, p, r)   x = A[r]   i = p-1   for j = p to r-1        do if A[j] <= x            then i = i+1                exchange(A[i], A[j])   exchange(A[i+1], A[r])   return i+1

算法解析

具体执行步骤如下：

i=p-1, j=p, x=A[r]
因为A[i]<=x，i=p，A[i]和A[j]交换
A[j]=7>x，j递增
当A[j]=1时，因为A[i]<=x，i=p+1，A[i]和A[j]交换
当A[j]=3时，因为A[i]<=x，i=p+2，A[i]和A[j]交换
A[j]=5>x，j递增
A[j]=6>x，j递增
j=r，A[r]和A[i+1]交换

源程序

#include <stdio.h>#include <unistd.h>#include <stdlib.h>#include <time.h> #include <sys/time.h>void swap(int &x, int &y); int partition(int *array, int left, int right); void qsort(int *array, int left, int right);void print(int *array, int left, int right);void print(int *array, int n);void input(int *&array, int n);const int number = 8;int main(int argc, char *argv[]){    //int array[] = {13, 19, 9, 5, 12, 8, 7, 4, 21, 2, 6, 11};    int array[] = {2, 8, 7, 1, 3, 5, 6, 4};    //int *array = new int[number];    int n = number;    //input(array, number);    printf("befor qsort\n");    print(array, n);    printf("\n");    struct timeval stv;    gettimeofday(&stv, NULL);    int st_usec = stv.tv_sec * 1000000 + stv.tv_usec;    qsort(array, 0, n - 1);    struct timeval end_tv;    gettimeofday(&end_tv, NULL);    int end_usec = end_tv.tv_sec * 1000000 + end_tv.tv_usec;    int cost = end_usec - st_usec;    printf("qsort cost time : %dus\n", cost);    printf("after qsort\n");    print(array, n);    //delete []array;    exit(0);}void qsort(int *array, int left, int right){    if (left < right)    {        //print(array, 0, number - 1);        int pos = partition(array, left, right);        qsort(array, left, pos - 1);        qsort(array, pos + 1, right);    }}int partition(int *array, int left, int right){    printf("before partition\n");    print(array, 0, number - 1);    int i = left - 1;    int x = array[right];    for (int j = left; j < right; j++)    {        printf("in partition : ");        if (array[j] <= x)        {            ++i;            swap(array[i], array[j]);        }        //print(array, 0, number - 1);        print(array, left, right);        //printf("in partition\n");    }    swap(array[i + 1], array[right]);    printf("after partition\n");    print(array, 0, number - 1);    printf("\n");    return i + 1;}void swap(int &x, int &y){    int temp = x;    x = y;    y = temp;}void print(int *array, int left, int right){    for (int i = left; i <= right; i++)    {        printf("%d ", array[i]);    }    printf("\n");}void print(int *array, int n){    print(array, 0, n - 1);}void input(int *&array, int n){    for (int i = 0; i < n; i++)    {        array[i] = rand() % n;    }}

算法执行过程

befor qsort2 8 7 1 3 5 6 4 before partition2 8 7 1 3 5 6 4 in partition : 2 8 7 1 3 5 6 4 in partition : 2 8 7 1 3 5 6 4 in partition : 2 8 7 1 3 5 6 4 in partition : 2 1 7 8 3 5 6 4 in partition : 2 1 3 8 7 5 6 4 in partition : 2 1 3 8 7 5 6 4 in partition : 2 1 3 8 7 5 6 4 after partition2 1 3 4 7 5 6 8 before partition2 1 3 4 7 5 6 8 in partition : 2 1 3 in partition : 2 1 3 after partition2 1 3 4 7 5 6 8 before partition2 1 3 4 7 5 6 8 in partition : 2 1 after partition1 2 3 4 7 5 6 8 before partition1 2 3 4 7 5 6 8 in partition : 7 5 6 8 in partition : 7 5 6 8 in partition : 7 5 6 8 after partition1 2 3 4 7 5 6 8 before partition1 2 3 4 7 5 6 8 in partition : 7 5 6 in partition : 5 7 6 after partition1 2 3 4 5 6 7 8 qsort cost time : 3610usafter qsort1 2 3 4 5 6 7 8

算法复杂度分析

时间复杂度空间复杂度说明最优情况 O(nlogn) O(logn) 平均情况 O(nlogn) O(logn) 最坏情况 O(n^2) O(n)

算法耗时分析

数量级耗时说明 1万 1.7ms 1百万 238ms 1千万 2.5s 2千万 5.4s 5千万 14s 1亿无内存空间有限

参考文献

[1] Thomas H.Cormen, Charles E.Leiserson,etc. 算法导论

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