用于排序数组中的天花板的 Python3 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，而下限是小于或等于 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

Python3

# Function to get index of ceiling of x in arr[low..high] */
def ceilSearch(arr, low, high, x):
  
    # 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 */
    i = low
    for i in range(high):
        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 and 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 */
arr = [1, 2, 8, 10, 10, 12, 19]
n = len(arr)
x = 3
index = ceilSearch(arr, 0, n-1, x);
  
if index == -1:
    print ("Ceiling of %d doesn't exist in array "% x)
else:
    print ("ceiling of %d is %d"%(x, arr[index]))
  
# This code is contributed by Shreyanshi Arun

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)。

Python3

# Function to get index of ceiling of x in arr[low..high]*/
def ceilSearch(arr, low, high, x):
  
    # 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] */ 
    elif arr[mid] < x:
        if mid + 1 <= high and 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 and x > arr[mid-1]:
            return mid
        else:
            return ceilSearch(arr, low, mid - 1, x)
   
# Driver program to check above functions */
arr = [1, 2, 8, 10, 10, 12, 19]
n = len(arr)
x = 20
index = ceilSearch(arr, 0, n-1, x);
  
if index == -1:
    print ("Ceiling of %d doesn't exist in array "% x)
else:
    print ("ceiling of %d is %d"%(x, arr[index]))
  
# This code is contributed by Shreyanshi Arun

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！