排序数组中天花板的 Java 程序
给定一个已排序的数组和值 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，假设数组按非递减顺序排序。写高效函数求楼层和天花板的 x. 例:
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，我们只实现了上限搜索功能。楼层搜索也可以用同样的方式实现。 方法 1(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)否则线性搜索索引 I，使 x 位于 arr[i]和 arr[i+1]之间。 3)如果在步骤 2 中没有找到索引 I，则返回-1
Java 语言(一种计算机语言，尤用于创建网站)
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(二分搜索法) 这里不用线性搜索，而是用二分搜索法来找出索引。二分搜索法将时间复杂度降低到 0(Logn)。
Java 语言(一种计算机语言，尤用于创建网站)
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)
相关文章: 排序数组中的 floor 在未排序数组中查找 Floor 和 ceil 如果您发现以上代码/算法中有任何一个不正确，或者找到更好的方法来解决相同的问题，或者想要为 Floor 实现共享代码，请写评论。
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