292.md

# 在相邻元素之间的差为 1 的数组中搜索元素

> 原文： [https://www.geeksforgeeks.org/search-an-element-in-an-array-where-difference-between-adjacent-elements-is-1/](https://www.geeksforgeeks.org/search-an-element-in-an-array-where-difference-between-adjacent-elements-is-1/)

给定一个数组，其中相邻元素之间的差为 1，编写一种算法来搜索数组中的元素并返回该元素的位置（返回第一个出现的元素）。

**示例**：

```
Let element to be searched be x

Input: arr[] = {8, 7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3}     
       x = 3
Output: Element 3 found at index 7

Input: arr[] =  {1, 2, 3, 4, 5, 4}
       x = 5
Output: Element 5 found at index 4

```

## [推荐：在继续进行解决之前，请先在 ***<u>{IDE}</u>*** 上尝试您的方法。](https://ide.geeksforgeeks.org/)

**简单方法**可以一次遍历给定数组，并将每个元素与给定元素“ x”进行比较。 如果匹配，则返回索引。

可以使用所有相邻元素之间的差异为 1 的事实对上述解决方案进行**优化。[想法]是从最左侧的元素开始进行比较，然后找出当前数组元素与 x 之间的差异。 将此差异设为“ diff”。 从 array 的给定属性中，我们始终知道 x 必须至少相距'diff'，因此，我们不逐一搜索，而是跳了'diff'。**

感谢 RajnishKrJha 建议此解决方案。

以下是上述想法的实现。

## C++ 

```

// C++ program to search an element in an array where 
// difference between all elements is 1 
#include<bits/stdc++.h> 
using namespace std; 

// x is the element to be searched in arr[0..n-1] 
int search(int arr[], int n, int x) 
{ 
    // Traverse the given array starting from 
    // leftmost element 
    int i = 0; 
    while (i<n) 
    { 
        // If x is found at index i 
        if (arr[i] == x) 
            return i; 

        // Jump the difference between current 
        // array element and x 
        i = i + abs(arr[i]-x); 
    } 

    cout << "number is not present!"; 
    return -1; 
} 

// Driver program to test above function 
int main() 
{ 
    int arr[] = {8 ,7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3 }; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    int x = 3; 
    cout << "Element " << x  << " is present at index "
         << search(arr,n,3); 
    return 0; 
} 

```

## 爪哇

```

// Java program to search an element in an 
// array where difference between all  
// elements is 1 

import java.io.*; 

class GFG { 

    // x is the element to be searched  
    // in arr[0..n-1] 
    static int search(int arr[], int n, int x) 
    { 

        // Traverse the given array starting  
        // from leftmost element 
        int i = 0; 
        while (i < n) 
        { 

            // If x is found at index i 
            if (arr[i] == x) 
                return i; 

            // Jump the difference between current 
            // array element and x 
            i = i + Math.abs(arr[i]-x); 
        } 

        System.out.println ("number is not" + 
                                     " present!"); 

        return -1; 
    } 

    // Driver program to test above function 
    public static void main (String[] args) { 

        int arr[] = {8 ,7, 6, 7, 6, 5, 4, 3,  
                                   2, 3, 4, 3 }; 
        int n = arr.length; 
        int x = 3; 
        System.out.println("Element " + x +  
                        " is present at index "
                            + search(arr,n,3)); 
    } 
} 

//This code is contributed by vt_m. 

```

## Python 3

```

# Python 3 program to search an element  
# in an array where difference between 
# all elements is 1 

# x is the element to be searched in  
# arr[0..n-1] 
def search(arr, n, x): 

    # Traverse the given array starting 
    # from leftmost element 
    i = 0
    while (i < n): 

        # If x is found at index i 
        if (arr[i] == x): 
            return i 

        # Jump the difference between 
        # current array element and x 
        i = i + abs(arr[i] - x) 

    print("number is not present!") 
    return -1

# Driver program to test above function 
arr = [8 ,7, 6, 7, 6, 5, 4, 3, 2, 3, 4, 3 ] 
n = len(arr) 
x = 3
print("Element" , x , " is present at index ", 
                             search(arr,n,3)) 

# This code is contributed by Smitha 

```

## C# 

```

// C# program to search an element  
// in an array where difference 
// between all elements is 1  
using System; 

public class GFG  
{ 

    // in arr[0..n-1] 
    static int search(int []arr, int n, 
                      int x) 
    { 

        // Traverse the given array starting  
        // from leftmost element 
        int i = 0; 
        while (i < n) 
        { 

            // If x is found at index i 
            if (arr[i] == x) 
                return i; 

            // Jump the difference between  
            // current array element and x 
            i = i + Math.Abs(arr[i] - x); 
        } 

        Console.WriteLine ("number is not" + 
                           " present!"); 

        return -1; 
    } 

    // Driver code 
    public static void Main()  
    { 

        int []arr = {8 ,7, 6, 7, 6, 5, 
                     4,3, 2, 3, 4, 3 }; 
        int n = arr.Length; 
        int x = 3; 
        Console.WriteLine("Element " + x +  
                        " is present at index "
                        + search(arr, n, 3)); 
    } 
} 

// This code is contributed by Sam007 

```

## PHP

```

<?php 
// PHP program to search an  
// element in an array where 
// difference between all  
// elements is 1 

// x is the element to be 
// searched in arr[0..n-1] 
function search($arr, $n, $x) 
{ 
    // Traverse the given array  
    // starting from leftmost 
    // element 
    $i = 0; 
    while ($i < $n) 
    { 
        // If x is found at index i 
        if ($arr[$i] == $x) 
            return $i; 

        // Jump the difference  
        // between current  
        // array element and x 
        $i = $i + abs($arr[$i] - $x); 
    } 

    echo "number is not present!"; 
    return -1; 
} 

// Driver Code 
$arr = array(8 ,7, 6, 7, 6, 5, 
             4, 3, 2, 3, 4, 3); 
$n = sizeof($arr); 
$x = 3; 
echo "Element " , $x , " is present " , 
      "at index ", search($arr, $n, 3); 

// This code is contributed 
// by nitin mittal. 
?> 

```

**输出**：

```
Element 3 is present at index 7
```

**[在相邻相差最多 k 的数组中搜索](https://www.geeksforgeeks.org/searching-array-adjacent-differ-k/)**

本文由 **Rishabh** 提供。 如果发现任何不正确的地方，或者想分享有关上述主题的更多信息，请发表评论

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