排序数组中天花板的 C# 程序
原文:https://www . geeksforgeeks . org/cs harp-program-for-child-in-a-sorted-array/
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
C
// C# program to find celing
// in a sorted array
using System;
class GFG {
// 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 code
public static void Main()
{
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)
Console.Write("Ceiling of " + x +
" doesn't exist in array");
else
Console.Write("ceiling of " + x +
" is " + arr[index]);
}
}
// This code is contributed by Sam007.
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)。
C
// C# program to find celing
// in a sorted array
using System;
class GFG {
// 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 code
public static void Main()
{
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)
Console.Write("Ceiling of " + x +
" doesn't exist in array");
else
Console.Write("ceiling of " + x +
" is " + arr[index]);
}
}
// This code is contributed by Sam007.
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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