c - 如何修改此快速排序算法的枢轴选择？

我想修改列出的快速排序算法，以选择枢轴选择为1）数组中的随机元素。 2）数组中第一个中间元素和最后一个元素的中位数。
====更新====
另外，如何实现一种计时机制来为每种算法计时？

void QuickSort1(long *L, int first, int last)
{
   int SplitPoint;  /* Value to Split list around */

   if( first < last )
   {
      /* Use first element as pivot (for splitting) */
      SplitPoint = first;                 /* assign SplitPoint to 1st index */
      Split(L, first, last, &SplitPoint); /* Split List around SplitPoint */
      QuickSort1(L, first, SplitPoint-1);  /* Sort 1st section of list */
      QuickSort1(L, SplitPoint+1, last);   /* Sort 2nd section of list */
   }
}

void Split(long *L, int first, int last, int *SplitPoint)
/* Splits a list around SplitPoint such that all values to the left of
   SplitPoint are < than SplitPoint and all values to the right of
   SplitPoint are >= SplitPoint */
{
   int x;            /* Key */
   int unknown;      /* index of unknown value */

   x = L[*SplitPoint];   /* assign x to value at SplitPoint */
   Swap( &L[*SplitPoint], &L[first] );
   *SplitPoint = first;

   /* Loop walks through unknown portion of list */
   for ( unknown = first+1; unknown <= last; ++unknown)
   {
      /* If unknown value is < SplitPoint Value, then: */
#ifdef TAKE_COUNT
         comparison_count++;
#endif
      if( L[unknown] < x ) {
         ++ (*SplitPoint);                     /* SplitPoint is incremented */
         Swap( &L[*SplitPoint], &L[unknown] ); /* values are swapped*/
      }
   }
   /* Original value which was being split upon is swapped with the current
      SplitPoint to put it in correct position */
   Swap( &L[first], &L[*SplitPoint] );
}

int FindMedian(long *L, int A, int B, int C)
/* Receives array and three respective indices in the array. */
/* Returns the index of the median. */
{
   if (L[A] < L[B])
     if (L[B] < L[C])       /*  A < B < C  */
       return (B);
     else
       if (L[A] < L[C])         /*  A < C < B  */
     return (C);
       else
     return (A);            /*  C < A < B  */
   else
     if (L[A] < L[C])           /*  B < A < C  */
       return (A);
     else
       if (L[B] < L[C])     /*  B < C < A  */
     return (C);
       else
     return (B);            /*  C < B < A  */
}

void Swap(long *a, long *b)
/* This function uses call by reference to switch two elements.*/
{
   long temp;    /* temporary variable used to switch. */

   temp = *a;   /* temp = 1st # */
   *a = *b;     /* 1st # = 2nd # */
   *b = temp;   /* 2nd # = temp */
}

如何针对列出的数据透视选项进行修改。还必须将它们添加为程序的其他菜单选项。

最佳答案

秘密在于，您始终选择数组中的第一个元素作为枢轴点。如果您希望使用另一个元素，这是一个很好的主意，如果输入已被排序或（通常情况下）已在排序时添加了一些新元素，则将其与第一个元素交换。现在，枢轴是第一个元素。

有了这种洞察力，任务应该变得非常简单。要选择随机枢轴，请生成N的随机数N-1，然后与第一个元素交换。取三位数的中位数，编写一组杂乱而简单明了的if ... else语句。