用于排序数组中的天花板的 C 程序

给定一个排序数组和一个值 x，x 的上限是数组中大于或等于 x 的最小元素，下限是小于或等于 x 的最大元素。假设数组按非降序排序。编写有效的函数来找到 x 的下限和上限。
例子 ：

For example, let the input array be {1, 2, 8, 10, 10, 12, 19}
For x = 0:    floor doesn't exist in array,  ceil  = 1
For x = 1:    floor  = 1,  ceil  = 1
For x = 5:    floor  = 2,  ceil  = 8
For x = 20:   floor  = 19,  ceil doesn't exist in array

在下面的方法中，我们只实现了天花板搜索功能。楼层搜索可以以相同的方式实现。
方法一（线性搜索）
搜索 x 上限的算法：
1) 如果 x 小于或等于数组中的第一个元素，则返回 0（第一个元素的索引）
2) Else 线性搜索索引 i，使得 x 位于 arr[i] 和 arr[i+1] 之间。
3) 如果我们在第 2 步中没有找到索引 i，则返回 -1

C

#include
  
/* Function to get index of ceiling of x in arr[low..high] */
int ceilSearch(int arr[], int low, int high, int x)
{
  int i;    
  
  /* If x is smaller than or equal to first element,
    then return the first element */
  if(x <= arr[low])
    return low;  
  
  /* Otherwise, linearly search for ceil value */
  for(i = low; i < high; i++)
  {
    if(arr[i] == x)
      return i;
  
    /* if x lies between arr[i] and arr[i+1] including
       arr[i+1], then return arr[i+1] */
    if(arr[i] < x && arr[i+1] >= x)
       return i+1;
  }         
  
  /* If we reach here then x is greater than the last element 
    of the array,  return -1 in this case */
  return -1;
}
  
  
/* Driver program to check above functions */
int main()
{
   int arr[] = {1, 2, 8, 10, 10, 12, 19};
   int n = sizeof(arr)/sizeof(arr[0]);
   int x = 3;
   int index = ceilSearch(arr, 0, n-1, x);
   if(index == -1)
     printf("Ceiling of %d doesn't exist in array ", x);
   else
     printf("ceiling of %d is %d", x, arr[index]);
   getchar();
   return 0;
}

输出 ：

ceiling of 3 is 8

时间复杂度： O(n)
方法 2（二分查找）
这里没有使用线性搜索，而是使用二进制搜索来查找索引。二分搜索将时间复杂度降低到 O(Logn)。

C

#include
  
/* Function to get index of ceiling of x in arr[low..high]*/
int ceilSearch(int arr[], int low, int high, int x)
{
  int mid;    
  
  /* If x is smaller than or equal to the first element,
    then return the first element */
  if(x <= arr[low])
    return low; 
  
  /* If x is greater than the last element, then return -1 */
  if(x > arr[high])
    return -1;  
  
  /* get the index of middle element of arr[low..high]*/
  mid = (low + high)/2;  /* low + (high - low)/2 */
  
  /* If x is same as middle element, then return mid */
  if(arr[mid] == x)
    return mid;
      
  /* If x is greater than arr[mid], then either arr[mid + 1]
    is ceiling of x or ceiling lies in arr[mid+1...high] */  
  else if(arr[mid] < x)
  {
    if(mid + 1 <= high && x <= arr[mid+1])
      return mid + 1;
    else 
      return ceilSearch(arr, mid+1, high, x);
  }
  
  /* If x is smaller than arr[mid], then either arr[mid] 
     is ceiling of x or ceiling lies in arr[low...mid-1] */    
  else
  {
    if(mid - 1 >= low && x > arr[mid-1])
      return mid;
    else     
      return ceilSearch(arr, low, mid - 1, x);
  }
}
  
/* Driver program to check above functions */
int main()
{
   int arr[] = {1, 2, 8, 10, 10, 12, 19};
   int n = sizeof(arr)/sizeof(arr[0]);
   int x = 20;
   int index = ceilSearch(arr, 0, n-1, x);
   if(index == -1)
     printf("Ceiling of %d doesn't exist in array ", x);
   else  
     printf("ceiling of %d is %d", x, arr[index]);
   getchar();
   return 0;
}

输出 ：

Ceiling of 20 doesn't exist in array 

时间复杂度：O(Logn)

有关更多详细信息，请参阅排序数组中有关天花板的完整文章！