未排序数组中的前后搜索

给定一个未排序的整数数组和一个元素x，请使用前后搜索查找x是否存在于数组中。
例子 ：

Input : arr[] = {10, 20, 80, 30, 60, 50, 
                     110, 100, 130, 170}
          x = 110;
Output : Yes

Input : arr[] = {10, 20, 80, 30, 60, 50, 
                     110, 100, 130, 170}
           x = 175;
Output : No

一个简单的解决方案是执行线性搜索。当要搜索的元素是数组中的最后一个元素时，线性搜索算法始终会导致O(n)的最坏情况复杂度。用于查找值为x的元素的前后搜索算法的工作方式如下：

1. 初始化索引front和back ，分别指向数组的第一个和最后一个元素。
2. 如果front大于Rear ，则返回false。
3. 检查前后索引处的元素x 。
4. 如果找到元素x，则返回true。
5. 否则递增前向和递减后向，然后转到步骤2。

关键点：

• 当元素在数组的中间或不存在时，最坏的情况是复杂度为O(n / 2)(等于O(n))。
• 当元素是数组中的第一个或最后一个元素时，最好的情况是O(1)。

C++

// CPP program to implement front and back
// search
#include
using namespace std;
 
bool search(int arr[], int n, int x)
{
    // Start searching from both ends
    int front = 0, back = n - 1;
 
    // Keep searching while two indexes
    // do not cross.
    while (front <= back)
    {
        if (arr[front] == x || arr[back] == x)
          return true;
        front++;
        back--;
    }
    return false;
}
 
int main()
{
   int arr[] = {10, 20, 80, 30, 60, 50,
                     110, 100, 130, 170};
   int x = 130;
   int n = sizeof(arr)/sizeof(arr[0]);
   if (search(arr, n, x))
      cout << "Yes";
   else
      cout << "No";
   return 0;
}

Java

// Java program to implement front and back
// search
class GFG {
     
    static boolean search(int arr[], int n, int x)
    {
         
        // Start searching from both ends
        int front = 0, back = n - 1;
     
        // Keep searching while two indexes
        // do not cross.
        while (front <= back)
        {
            if (arr[front] == x || arr[back] == x)
                return true;
            front++;
            back--;
        }
         
        return false;
    }
     
    // driver code
    public static void main (String[] args)
    {
        int arr[] = {10, 20, 80, 30, 60, 50,
                        110, 100, 130, 170};
        int x = 130;
        int n = arr.length;
         
        if (search(arr, n, x))
            System.out.print("Yes");
        else
            System.out.print("No");
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python program to implement
# front and back search
 
def search(arr, n, x):
 
    # Start searching from both ends
    front = 0; back = n - 1
     
    # Keep searching while two
    # indexes do not cross.
    while (front <= back):
     
        if (arr[front] == x or arr[back] == x):
            return True
        front += 1
        back -= 1
     
    return False
 
# Driver code
arr = [10, 20, 80, 30, 60,
       50, 110, 100, 130, 170]
x = 130
n = len(arr)
 
if (search(arr, n, x)):
    print("Yes")
else:
    print("No")
     
# This code is contributed by Anant Agarwal.

C#

// C# program to implement front and back
// search
using System;
 
public class GFG {
 
    static bool search(int []arr, int n, int x)
    {
         
        // Start searching from both ends
        int front = 0, back = n - 1;
     
        // Keep searching while two indexes
        // do not cross.
        while (front <= back)
        {
            if (arr[front] == x || arr[back] == x)
                return true;
            front++;
            back--;
        }
         
        return false;
    }
 
    static public void Main ()
    {
        int []arr = {10, 20, 80, 30, 60, 50,
                    110, 100, 130, 170};
        int x = 130;
        int n = arr.Length;
         
        if (search(arr, n, x))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by vt_m.

Javascript

 
// JavaScript program to implement front and back
 
    function search(arr, n, x)
    {
           
        // Start searching from both ends
        let front = 0, back = n - 1;
       
        // Keep searching while two indexes
        // do not cross.
        while (front <= back)
        {
            if (arr[front] == x || arr[back] == x)
                return true;
            front++;
            back--;
        }
           
        return false;
    }
 
// Driver Code
 
        let arr = [10, 20, 80, 30, 60, 50,
                        110, 100, 130, 170];
        let x = 130;
        let n = arr.length;
           
        if (search(arr, n, x))
            document.write("Yes");
        else
            document.write("No");
 
// This code is contributed by avijitmondal1998.

输出：

Yes