用于排序数组中天花板的 Javascript 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

Javascript

  
/* Function to get index of ceiling of 
x in arr[low..high] */
function ceilSearch(arr, low, high, x) 
{ 
      
    let 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 code 
  
    let arr = [1, 2, 8, 10, 10, 12, 19]; 
    let n = arr.length; 
    let x = 3; 
    let index = ceilSearch(arr, 0, n-1, x); 
    if(index == -1) 
        document.write("Ceiling of " + x + " doesn't exist in array "); 
    else
        document.write ("ceiling of " + x + " is " + arr[index]);  
  
  

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)。

Javascript

// Javascript Program for Ceiling in  
// a sorted array 
    
// Function to get index of ceiling 
// of x in arr[low..high] 
function ceilSearch(arr, low, high, x) 
{ 
    let 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] */
    // low + (high - low)/2 
    mid = (low + high)/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 Code 
let arr = [1, 2, 8, 10, 10, 12, 19]; 
let n = arr.length; 
let x = 20; 
let index = ceilSearch(arr, 0, n - 1, x); 
  
if(index == -1){
    document.write(`Ceiling of ${x} doesn't exist in array `); 
}else{
    document.write(`ceiling of ${x} is ${arr[index]}`);  
}
    
// This code is contributed by _saurabh_jaiswal. 
  

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！