用于排序数组中的天花板的Java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

Java

class Main
{
    /* Function to get index of ceiling 
       of x in arr[low..high] */
    static 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 */
    public static void main (String[] args)
    {
       int arr[] = {1, 2, 8, 10, 10, 12, 19};
       int n = arr.length;
       int x = 3;
       int index = ceilSearch(arr, 0, n-1, x);
       if(index == -1)
         System.out.println("Ceiling of "+x+" doesn't exist in array");
       else
         System.out.println("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)。

Java

class Main
{
    /* Function to get index of 
       ceiling of x in arr[low..high]*/
    static 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 */
    public static void main (String[] args)
    {
       int arr[] = {1, 2, 8, 10, 10, 12, 19};
       int n = arr.length;
       int x = 8;
       int index = ceilSearch(arr, 0, n-1, x);
       if(index == -1)
         System.out.println("Ceiling of "+x+" doesn't exist in array");
       else 
         System.out.println("ceiling of "+x+" is "+arr[index]);
    }  
}

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！