快速排序比较计数间歇性结果

Question

SOLUTION 该解决方案对我的代码来说是独一无二的——我将 srand(time(NULL)); 放在了循环中，而它本应放在循环外

我正在尝试计算快速排序算法中的比较次数。我有一个递归版本工作正常，但它一直出现段错误，因为我使用的是大数组大小——运行 out of stack space.

所以现在我得出了一种迭代方法，并且它有效。也就是说，除了我的比较次数计数器。

它正在返回间歇性结果，例如...

unsorted: [9][8][7][6][5][4][3][2][1][0]
sorted: [0][1][2][3][4][5][6][7][8][9]
Numer of comparisons: 22

unsorted: [9][8][7][6][5][4][3][2][1][0]
sorted: [0][1][2][3][4][5][6][7][8][9]
Numer of comparisons: 19749794

unsorted: [9][8][7][6][5][4][3][2][1][0]
sorted: [0][1][2][3][4][5][6][7][8][9]
Numer of comparisons: 6088231

我的迭代快速排序代码是...

#include <time.h>
#define BUFLEN 6400

extern int buf[BUFLEN];
extern int quick_count; //comparison count

struct stack {
    int stk[BUFLEN];
    int top;
};

struct stack s;

void push(int x);
int pop();

void iterative_quick_sort (int buf[], int n) {
    int left_ptr, right_ptr, pivot_index, pivot, temp, l, r;
    if (n < 2) //case the partitioning has reached the atomic element
        return;
    r = n - 1;
    l = 0;
    s.top = -1;
    loop: do{
      srand(time(NULL));
      if ((r - l) == 0)
        pivot_index = 1;
      else {
        pivot_index = rand() % (r - l);
        pivot_index += l;
      }
      pivot = buf[pivot_index]; //pivot holds the value of the pivot element
      left_ptr = l;
      right_ptr = r;
      if ((r - l) != 0 || (r - l) != 1){
        while (1) {
            while (buf[left_ptr] < pivot){ //loop and increment left_ptr until an element on the left side is larger than the pivot
              left_ptr++;
            } //now left_ptr holds the index of the value that needs to be swapped with an element from the right side
            while (pivot < buf[right_ptr]){ //loop and increment right_ptr until an element on the right side is smaller than the pivot
              right_ptr--;
            } //now right_ptr holds the index of the value that needs to be swapped with an element from the left side
            quick_count++;
            if (left_ptr >= right_ptr)
                break; //once the pivots reach or overlap each other, break the loop
            //perform swap with temporary variable temp
            temp = buf[left_ptr];
            buf[left_ptr] = buf[right_ptr];
            buf[right_ptr] = temp;
        }
      }

      if (l == (n - 2))
        break;
      else if ((r - l) >= 2){
        //goto loop with left side values
        push(r);
        r = pivot_index + 1;
        goto loop;
      }
      else {
        //goto loop with right side values
        l = r;
        r = pop();
        goto loop;
      }
  }while(1);
}

//cite http://www.sanfoundry.com/c-program-stack-implementation/
void push (int x){
    s.top = s.top + 1;
    s.stk[s.top] = x;
}

int pop(){
    int x = s.stk[s.top];
    s.top = s.top - 1;
    return x;
}

根据请求，我添加了调用快速排序的函数（注意：quick_count 作为全局变量初始化为零——用作外部变量）

int unsorted_quick[] = {9,8,7,6,5,4,3,2,1,0}; //n = 10
//print unsorted_quick
  printf("\nSecond, we sort the following array by using the quick sort algorithm\n");
  for (i = 0; i < 10; i++){
    printf("[%d]", unsorted_quick[i]);
  }
  printf("\n");

  //fill buf with the unsorted quick array
  for (i = 0; i < 10; i++){
    buf[i] = unsorted_quick[i];
  }

  iterative_quick_sort(buf, 10); //call quick_sort()

  //print sorted
  for (i = 0; i < 10; i++){
    printf("[%d]", buf[i]);
  }
  printf("\nNumber of comparisons: %d\n", quick_count);  //print count

Answer 1

您在选择随机枢轴的循环内调用 srand(time(NULL))。此函数必须调用一次以初始化随机数生成器的状态。生成器需要一个通过调用 srand() 设置的起始种子。然后，给定种子，对 rand() 的每次后续调用都会以可重现的序列为您提供一个随机数。从相同的种子开始，您将获得相同的随机序列。

问题是您在循环中设置了种子并且种子是相同的数字，因此您将始终获得相同的 "random" 值。发生这种情况是因为 time(NULL) 是以秒为单位从当前时间获取的，这意味着随机数在同一秒内是相同的。

你必须把它放在循环之前：do {

这里对正在发生的事情有一个很好的解释：Problems when calling srand(time(NULL)) inside rollDice function

还有这里：srand() — why call it only once?

快速排序比较计数间歇性结果

quicksort comparison count intermittent results

c

count

comparison

iteration

quicksort