给定一个由n个整数组成的数组。每个数组元素都是通过将+1或-1加到上一个元素而获得的，即，任意两个连续元素之间的绝对差为1。任务是以最小的比较次数搜索元素索引(少于逐个元素进行的简单搜索) 。如果该元素多次出现，则打印最小的索引。如果不存在该元素，则打印-1。
例子：
Input : arr[] = {5, 4, 5, 6, 5, 4, 3, 2}
x = 4.
Output : 1
The first occurrence of element x is at
index 1.
Input : arr[] = { 5, 4, 5, 6, 4, 3, 2, 3 }
x = 9.
Output : -1
Element x is not present in arr[]令要搜索的元素是x。在任何索引i处，如果arr [i]！= x，则x出现的可能性在位置i + abs(arr [i] – a)处，因为每个元素都是通过将+1或-1加到上一个元素。在i和i + abs(arr [i] – a)之间不可能存在el。因此，如果arr [i]！= x，则直接跳转到i + abs(arr [i] – a)。
Algorithm to solve the problem:
1. Start from index = 0.
2. Compare arr[index] and x.
a) If both are equal, return index.
b) If not, set index = index + abs(arr[index] - x).
3. Repeat step 2.下面是上述想法的实现：
C++
// C++ program to search an element in an array
// where each element is obtained by adding
// either +1 or -1 to previous element.
#include
using namespace std;
// Return the index of the element to be searched.
int search(int arr[], int n, int x)
{
// Searching x in arr[0..n-1]
int i = 0;
while (i <= n-1)
{
// Checking if element is found.
if (arr[i] == x)
return i;
// Else jumping to abs(arr[i]-x).
i += abs(arr[i]-x);
}
return -1;
}
// Driven Program
int main()
{
int arr[] = {5, 4, 5, 6, 5, 4, 3, 2};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 4;
cout << search(arr, n, x) << endl;
return 0;
} Java
// Java program to search an element
// in an array where each element is
// obtained by adding either +1 or
// -1 to previous element.
class GFG
{
// Return the index of the
// element to be searched.
static int search(int arr[], int n, int x)
{
// Searching x in arr[0..n-1]
int i = 0;
while (i <= n-1)
{
// Checking if element is found.
if (arr[i] == x)
return i;
// Else jumping to abs(arr[i]-x).
i += Math.abs(arr[i]-x);
}
return -1;
}
// Driver code
public static void main (String[] args)
{
int arr[] = {5, 4, 5, 6, 5, 4, 3, 2};
int n = arr.length;
int x = 4;
System.out.println(search(arr, n, x));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python program to search an element in
# an array where each element is obtained
# by adding either +1 or -1 to previous element
# Return the index of the element to be searched
def search(arr, n, x):
# Searching x in arr[0..n-1]
i = 0
while (i <= n-1):
# Checking if element is found.
if (arr[i] == x):
return i
# Else jumping to abs(arr[i]-x).
i += abs(arr[i] - x)
return -1
# Driver code
arr = [5, 4, 5, 6, 5, 4, 3, 2]
n = len(arr)
x = 4
print(search(arr, n, x))
# This code is contributed by Anant Agarwal.C#
// C# program to search an element in
// an array where each element is
// obtained by adding either + 1 or
// -1 to previous element.
using System;
class GFG
{
// Return the index of the
// element to be searched.
static int search(int []arr, int n,
int x)
{
// Searching x in arr[0.. n - 1]
int i = 0;
while (i <= n - 1)
{
// Checking if element is found
if (arr[i] == x)
return i;
// Else jumping to abs(arr[i] - x)
i += Math.Abs(arr[i] - x);
}
return -1;
}
// Driver code
public static void Main ()
{
int []arr = {5, 4, 5, 6, 5, 4, 3, 2};
int n = arr.Length;
int x = 4;
Console.WriteLine(search(arr, n, x));
}
}
// This code is contributed by vt_m.PHP
Javascript
输出：
1