排序数组中的地板

给定一个已排序的数组和一个值 x，x 的下限是数组中小于或等于 x 的最大元素。编写有效的函数来找到 x 的下限。
例子：

Input : arr[] = {1, 2, 8, 10, 10, 12, 19}, x = 5
Output : 2
2 is the largest element in 
arr[] smaller than 5.

Input : arr[] = {1, 2, 8, 10, 10, 12, 19}, x = 20
Output : 19
19 is the largest element in
arr[] smaller than 20.

Input : arr[] = {1, 2, 8, 10, 10, 12, 19}, x = 0
Output : -1
Since floor doesn't exist,
output is -1.

简单方法
做法：思路很简单，遍历数组，找到第一个大于x的元素。刚找到元素之前的元素是 x 的下限。
算法：

1. 从头到尾遍历数组。
2. 如果当前元素大于 x 打印前一个数字并中断循环。
3. 如果没有大于 x 的数字，则打印最后一个元素
4. 如果第一个数字大于 x 则打印 -1

C++

// C/C++ program to find floor of a given number
// in a sorted array
#include 
 
/* An inefficient function to get
index of floor of x in arr[0..n-1] */
int floorSearch(int arr[], int n, int x)
{
    // If last element is smaller than x
    if (x >= arr[n - 1])
        return n - 1;
 
    // If first element is greater than x
    if (x < arr[0])
        return -1;
 
    // Linearly search for the first element
    // greater than x
    for (int i = 1; i < n; i++)
        if (arr[i] > x)
            return (i - 1);
 
    return -1;
}
 
/* Driver program to check above functions */
int main()
{
    int arr[] = { 1, 2, 4, 6, 10, 12, 14 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 7;
    int index = floorSearch(arr, n - 1, x);
    if (index == -1)
        printf("Floor of %d doesn't exist in array ", x);
    else
        printf("Floor of %d is %d", x, arr[index]);
    return 0;
}

Java

// Java program to find floor of
// a given number in a sorted array
import java.io.*;
import java.util.*;
import java.lang.*;
 
class GFG {
 
    /* An inefficient function to get index of floor
of x in arr[0..n-1] */
    static int floorSearch(
        int arr[], int n, int x)
    {
        // If last element is smaller than x
        if (x >= arr[n - 1])
            return n - 1;
 
        // If first element is greater than x
        if (x < arr[0])
            return -1;
 
        // Linearly search for the first element
        // greater than x
        for (int i = 1; i < n; i++)
            if (arr[i] > x)
                return (i - 1);
 
        return -1;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 4, 6, 10, 12, 14 };
        int n = arr.length;
        int x = 7;
        int index = floorSearch(arr, n - 1, x);
        if (index == -1)
            System.out.print(
                "Floor of " + x
                + " doesn't exist in array ");
        else
            System.out.print(
                "Floor of " + x + " is "
                + arr[index]);
    }
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Python3

# Python3 program to find floor of a
# given number in a sorted array
 
# Function to get index of floor
# of x in arr[low..high]
def floorSearch(arr, low, high, x):
 
    # If low and high cross each other
    if (low > high):
        return -1
 
    # If last element is smaller than x
    if (x >= arr[high]):
        return high
 
    # Find the middle point
    mid = int((low + high) / 2)
 
    # If middle point is floor.
    if (arr[mid] == x):
        return mid
 
    # If x lies between mid-1 and mid
    if (mid > 0 and arr[mid-1] <= x
                and x < arr[mid]):
        return mid - 1
 
    # If x is smaller than mid,
    # floor must be in left half.
    if (x < arr[mid]):
        return floorSearch(arr, low, mid-1, x)
 
    # If mid-1 is not floor and x is greater than
    # arr[mid],
    return floorSearch(arr, mid + 1, high, x)
 
 
# Driver Code
arr = [1, 2, 4, 6, 10, 12, 14]
n = len(arr)
x = 7
index = floorSearch(arr, 0, n-1, x)
 
if (index == -1):
    print("Floor of", x, "doesn't exist \
                    in array ", end = "")
else:
    print("Floor of", x, "is", arr[index])
 
# This code is contributed by Smitha Dinesh Semwal.

C#

// C# program to find floor of a given number
// in a sorted array
using System;
 
class GFG {
 
    /* An inefficient function to get index of floor
of x in arr[0..n-1] */
    static int floorSearch(int[] arr, int n, int x)
    {
        // If last element is smaller than x
        if (x >= arr[n - 1])
            return n - 1;
 
        // If first element is greater than x
        if (x < arr[0])
            return -1;
 
        // Linearly search for the first element
        // greater than x
        for (int i = 1; i < n; i++)
            if (arr[i] > x)
                return (i - 1);
 
        return -1;
    }
 
    // Driver Code
    static void Main()
    {
        int[] arr = { 1, 2, 4, 6, 10, 12, 14 };
        int n = arr.Length;
        int x = 7;
        int index = floorSearch(arr, n - 1, x);
        if (index == -1)
            Console.WriteLine("Floor of " + x + " doesn't exist in array ");
        else
            Console.WriteLine("Floor of " + x + " is " + arr[index]);
    }
}
 
// This code is contributed
// by mits

PHP

= $arr[$n - 1])
        return $n - 1;
 
    // If first element is greater
    // than x
    if ($x < $arr[0])
        return -1;
 
    // Linearly search for the
    // first element greater than x
    for ($i = 1; $i < $n; $i++)
    if ($arr[$i] > $x)
        return ($i - 1);
 
    return -1;
}
 
// Driver Code
$arr = array (1, 2, 4, 6, 10, 12, 14);
$n = sizeof($arr);
$x = 7;
$index = floorSearch($arr, $n - 1, $x);
if ($index == -1)
    echo "Floor of ", $x,
         "doesn't exist in array ";
else
    echo "Floor of ", $x,
         " is ", $arr[$index];
     
// This code is contributed by ajit
?>

Javascript

 
// JavaScript program to find floor of
// a given number in a sorted array
 
    /* An inefficient function to get index of floor
of x in arr[0..n-1] */
    function floorSearch(arr, n, x)
    {
        // If last element is smaller than x
        if (x >= arr[n - 1])
            return n - 1;
   
        // If first element is greater than x
        if (x < arr[0])
            return -1;
   
        // Linearly search for the first element
        // greater than x
        for (let i = 1; i < n; i++)
            if (arr[i] > x)
                return (i - 1);
   
        return -1;
    }
 
// Driver Code
 
        let arr = [ 1, 2, 4, 6, 10, 12, 14 ];
        let n = arr.length;
        let x = 7;
        let index = floorSearch(arr, n - 1, x);
        if (index == -1)
             document.write(
                "Floor of " + x
                + " doesn't exist in array ");
        else
             document.write(
                "Floor of " + x + " is "
                + arr[index]);
                         

输出：

Floor of 7 is 6.

复杂度分析：

• 时间复杂度： O(n)。
  遍历一个数组只需要一个循环，所以时间复杂度是 O(n)。
• 空间复杂度： O(1)。
  不需要额外的空间，所以空间复杂度是恒定的

高效的方法
方法：问题中有一个问题，给定的数组已排序。这个想法是使用二分搜索通过将数字x与中间元素进行比较并将搜索空间分成两半来找到排序数组中数字x的下限。
算法：

1. 该算法可以通过递归或迭代实现，但基本思想保持不变。
2. 有基本情况需要处理。
   1. 如果没有大于 x 的数字，则打印最后一个元素
   2. 如果第一个数字大于 x 则打印 -1
3. 创建三个变量low = 0 ， mid 和high = n-1和另一个变量来存储答案
4. 运行循环或递归直到并且除非低小于或等于高。
5. 检查中间 ( (low + high) /2 ) 元素是否小于 x，如果是，则更新低，即low = mid + 1 ，并使用中间元素更新答案。在这一步中，我们将搜索空间减少到一半。
6. 否则更新低，即高 = 中 – 1
7. 打印答案。

C++

// A C/C++ program to find floor
// of a given number in a sorted array
#include 
 
/* Function to get index of floor of x in
   arr[low..high] */
int floorSearch(int arr[], int low,
                int high, int x)
{
    // If low and high cross each other
    if (low > high)
        return -1;
 
    // If last element is smaller than x
    if (x >= arr[high])
        return high;
 
    // Find the middle point
    int mid = (low + high) / 2;
 
    // If middle point is floor.
    if (arr[mid] == x)
        return mid;
 
    // If x lies between mid-1 and mid
    if (mid > 0 && arr[mid - 1] <= x
        && x < arr[mid])
        return mid - 1;
 
    // If x is smaller than mid, floor
    // must be in left half.
    if (x < arr[mid])
        return floorSearch(
            arr, low, mid - 1, x);
 
    // If mid-1 is not floor and x is
    // greater than arr[mid],
    return floorSearch(arr, mid + 1, high, x);
}
 
/* Driver program to check above functions */
int main()
{
    int arr[] = { 1, 2, 4, 6, 10, 12, 14 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 7;
    int index = floorSearch(arr, 0, n - 1, x);
    if (index == -1)
        printf(
            "Floor of %d doesn't exist in array ", x);
    else
        printf(
            "Floor of %d is %d", x, arr[index]);
    return 0;
}

Java

// Java program to find floor of
// a given number in a sorted array
import java.io.*;
 
class GFG {
 
    /* Function to get index of floor of x in
    arr[low..high] */
    static int floorSearch(
        int arr[], int low,
        int high, int x)
    {
        // If low and high cross each other
        if (low > high)
            return -1;
 
        // If last element is smaller than x
        if (x >= arr[high])
            return high;
 
        // Find the middle point
        int mid = (low + high) / 2;
 
        // If middle point is floor.
        if (arr[mid] == x)
            return mid;
 
        // If x lies between mid-1 and mid
        if (
            mid > 0 && arr[mid - 1] <= x && x < arr[mid])
            return mid - 1;
 
        // If x is smaller than mid, floor
        // must be in left half.
        if (x < arr[mid])
            return floorSearch(
                arr, low,
                mid - 1, x);
 
        // If mid-1 is not floor and x is
        // greater than arr[mid],
        return floorSearch(
            arr, mid + 1, high,
            x);
    }
 
    /* Driver program to check above functions */
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 4, 6, 10, 12, 14 };
        int n = arr.length;
        int x = 7;
        int index = floorSearch(
            arr, 0, n - 1,
            x);
        if (index == -1)
            System.out.println(
                "Floor of " + x + " dosen't exist in array ");
        else
            System.out.println(
                "Floor of " + x + " is " + arr[index]);
    }
}
// This code is contributed by Prerna Saini

蟒蛇3

# Python3 program to find floor of a 
# given number in a sorted array
 
# Function to get index of floor
# of x in arr[low..high]
def floorSearch(arr, low, high, x):
 
    # If low and high cross each other
    if (low > high):
        return -1
 
    # If last element is smaller than x
    if (x >= arr[high]):
        return high
 
    # Find the middle point
    mid = int((low + high) / 2)
 
    # If middle point is floor.
    if (arr[mid] == x):
        return mid
 
    # If x lies between mid-1 and mid
    if (mid > 0 and arr[mid-1] <= x
                and x < arr[mid]):
        return mid - 1
 
    # If x is smaller than mid, 
    # floor must be in left half.
    if (x < arr[mid]):
        return floorSearch(arr, low, mid-1, x)
 
    # If mid-1 is not floor and x is greater than
    # arr[mid],
    return floorSearch(arr, mid + 1, high, x)
 
 
# Driver Code
arr = [1, 2, 4, 6, 10, 12, 14]
n = len(arr)
x = 7
index = floorSearch(arr, 0, n-1, x)
 
if (index == -1):
    print("Floor of", x, "doesn't exist\
                    in array ", end = "")
else:
    print("Floor of", x, "is", arr[index])
 
# This code is contributed by Smitha Dinesh Semwal.

C＃

// C# program to find floor of
// a given number in a sorted array
using System;
 
class GFG {
 
    /* Function to get index of floor of x in
    arr[low..high] */
    static int floorSearch(
        int[] arr, int low,
        int high, int x)
    {
 
        // If low and high cross each other
        if (low > high)
            return -1;
 
        // If last element is smaller than x
        if (x >= arr[high])
            return high;
 
        // Find the middle point
        int mid = (low + high) / 2;
 
        // If middle point is floor.
        if (arr[mid] == x)
            return mid;
 
        // If x lies between mid-1 and mid
        if (mid > 0 && arr[mid - 1] <= x && x < arr[mid])
            return mid - 1;
 
        // If x is smaller than mid, floor
        // must be in left half.
        if (x < arr[mid])
            return floorSearch(arr, low,
                               mid - 1, x);
 
        // If mid-1 is not floor and x is
        // greater than arr[mid],
        return floorSearch(arr, mid + 1, high,
                           x);
    }
 
    /* Driver program to check above functions */
    public static void Main()
    {
        int[] arr = { 1, 2, 4, 6, 10, 12, 14 };
        int n = arr.Length;
        int x = 7;
        int index = floorSearch(arr, 0, n - 1,
                                x);
        if (index == -1)
            Console.Write("Floor of " + x + " dosen't exist in array ");
        else
            Console.Write("Floor of " + x + " is " + arr[index]);
    }
}
 
// This code is contributed by nitin mittal.

Javascript

 
// javascript program to find floor of
// a given number in a sorted array
 
/* Function to get index of floor of x in
arr[low..high] */
function floorSearch(
    arr , low,
    high , x)
{
    // If low and high cross each other
    if (low > high)
        return -1;
 
    // If last element is smaller than x
    if (x >= arr[high])
        return high;
 
    // Find the middle point
    var mid = (low + high) / 2;
 
    // If middle povar is floor.
    if (arr[mid] == x)
        return mid;
 
    // If x lies between mid-1 and mid
    if (
        mid > 0 && arr[mid - 1] <= x && x < arr[mid])
        return mid - 1;
 
    // If x is smaller than mid, floor
    // must be in left half.
    if (x < arr[mid])
        return floorSearch(
            arr, low,
            mid - 1, x);
 
    // If mid-1 is not floor and x is
    // greater than arr[mid],
    return floorSearch(
        arr, mid + 1, high,
        x);
}
 
/* Driver program to check above functions */
var arr = [ 1, 2, 4, 6, 10, 12, 14 ];
var n = arr.length;
var x = 7;
var index = floorSearch(
    arr, 0, n - 1,
    x);
if (index == -1)
    document.write(
        "Floor of " + x + " dosen't exist in array ");
else
    document.write(
        "Floor of " + x + " is " + arr[index]);
 
// This code is contributed by Amit Katiyar

输出：

Floor of 7 is 6.

复杂度分析：

• 时间复杂度： O(log n)。
  要运行二进制搜索，所需的时间复杂度为 O(log n)。
• 空间复杂度： O(1)。
  由于不需要额外的空间，所以空间复杂度是恒定的。

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