# 天花板排列 > 原文： [https://www.geeksforgeeks.org/ceiling-in-a-sorted-array/](https://www.geeksforgeeks.org/ceiling-in-a-sorted-array/) 給定一個排序數組和一個值 x，x 的上限是數組中大于或等于 x 的最小元素，下限是小于或等于 x 的最大元素。 假設數組以非降序排序。 編寫有效的函數以查找 x 的底和頂。 **示例**： ``` 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 ``` 在以下方法中，我們僅實現了上限搜索功能。 樓層搜索可以以相同的方式實現。 **方法 1（線性搜索）** 用于搜索 x 上限的算法： 1）如果 x 小于或等于數組中的第一個元素，則返回 0（第一個元素的索引）[ 2）其他線性搜索索引 i，以使 x 介于 arr [i]和 arr [i + 1]之間。 3）如果在步驟 2 中找不到索引 i，則返回-1 ## C++ ```cpp // C++ implementation of above approach #include <bits/stdc++.h> using namespace std; /* Function to get index of ceiling of x in arr[low..high] */ 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*/ int main()? {? ????int arr[] = {1, 2, 8, 10, 10, 12, 19};? ????int n = sizeof(arr)/sizeof(arr[0]);? ????int x = 3;? ????int index = ceilSearch(arr, 0, n-1, x);? ????if(index == -1)? ????????cout << "Ceiling of " << x << " doesn't exist in array ";? ????else ????????cout << "ceiling of " << x << " is " << arr[index];? ????return 0;? }? // This is code is contributed by rathbhupendra ```