Question

我尝试在C ++中实现上述算法，但是在计算数组的一半长度时，我得到一个舍入错误。

尝试：

#include <iostream>
#include <math.h>
using namespace std;

void print_array(int arr[], int size){
    for(int i = 0; i < size; i++){
        cout << arr[i] << "-";
    }
    cout << endl;
}

float median_h(int arr[], int size){
  float median = 0.0;
  if(size%2 == 0)
    median = (arr[size/2] + arr[(size/2)-1])/2;
  else median = arr[(int)ceil(size/2)];
  return median;
}

float median(int arr1[], int arr2[], int size_1, int size_2){
   cout << size_1 << endl;
   print_array(arr1,size_1);
   cout << size_2 << endl;
   print_array(arr2,size_2);
   cout << endl;
   float m1 = median_h(arr1,size_1);
   float m2 = median_h(arr2,size_2);
   float median_res = 0.0;
   if(size_1 == 1)
      median_res = (arr1[0] + arr2[0])/2;
   else if(m1 == m2)
      median_res = m1;
   else{
     int index = 0;
     int size = ceil(size_1/2);
     (size_1 % 2 == 0)? index = size_1/2 : index = floor(size_1/2); 
     if(m1 < m2)
       median_res = median(arr1 + index,arr2, size, size);    
     else
       median_res = median(arr1,arr2 + index, size, size);
   }
   return median_res;
}



int main(void){
    int arr1[] = {1,12,15,26,38};
    int arr2[] = {2,13,17,30,45};

    float med = median(arr1,arr2,5,5);
    cout << med << endl;

}

这是输出：

  output:

    5
    1-12-15-26-38-
    5
    2-13-17-30-45-

    2
    15-26-
    2
    2-13-

    1
    15-
    1
    13-

    14

我希望在第二次递归迭代中长度为3，但我得到2.我不知道ceil（5/2）的错误。它应该是3。

Answer 1

 int size = ceil(size_1/2);

由于size_1的类型为int，因此将其除以2隐含等效于取得除法的最低限度。例如如果size_1等于3，则（size_1 / 2）将返回1，而不是1.5。

因此，ceil（）不会在这里得到你想要的东西，因为传递给它的值已经被覆盖了。

你可以这样做：

int size = ceil(size_1/2.0f);  // now we're dividing by a float, so the result will be non-integer

...但你可以得到相同的结果，而不必仅仅通过在分割前将值加1来求助于浮点数学：

int size = (size_1+1)/2;

Answer 2

/* Function to get median of a sorted array */
int median_h(int arr[], int n)
{
    if (n%2 == 0)
        return (arr[n/2] + arr[n/2-1])/2;
    else
        return arr[n/2]; // <- HERE
    // also since n is int, you can do ( n << 1 )
    // it's faster
}

/* This function returns median of ar1[] and ar2[].
   Assumptions in this function:
   Both ar1[] and ar2[] are sorted arrays
   Both have n elements */
int median(int ar1[], int ar2[], int n)
{
    /* return -1  for invalid input */
    if (n <= 0)
        return -1;
    if (n == 1)
        return (ar1[0] + ar2[0])/2;
    if (n == 2)
        return (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1])) / 2;

    int m1 = median_h(ar1, n); /* get the median of the first array */
    int m2 = median_h(ar2, n); /* get the median of the second array */

    /* If medians are equal then return either m1 or m2 */
    if (m1 == m2)
        return m1;

    /* if m1 < m2 then median must exist in ar1[m1....] and
        ar2[....m2] */
    if (m1 < m2)
    {
        if (n % 2 == 0)
            return median(ar1 + n/2 - 1, ar2, n - n/2 +1);
        return median(ar1 + n/2, ar2, n - n/2);
    }

    /* if m1 > m2 then median must exist in ar1[....m1] and
        ar2[m2...] */
    if (n % 2 == 0)
        return median(ar2 + n/2 - 1, ar1, n - n/2 + 1);
    return median(ar2 + n/2, ar1, n - n/2);
}

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在C ++中实现两个大小相等的排序数组的中值

2 个答案: