相邻元素之间差异为 0 或 1 的最大长度子数组

给定一个包含n 个整数的数组。任务是找到子数组的最大长度，使得子数组的所有连续元素之间的绝对差为0或1 。
例子：

Input: arr[] = {2, 5, 6, 3, 7, 6, 5, 8}
Output: 3
{5, 6} and {7, 6, 5} are the only valid sub-arrays.
Input: arr[] = {-2, -1, 5, -1, 4, 0, 3}
Output: 2

编程需要懂一点英语

方法：从数组的第一个元素开始，找到第一个有效的子数组并存储它的长度，然后从下一个元素（第一个未包含在第一个子数组中的元素）开始，找到另一个有效的子数组大批。重复该过程，直到找到所有有效子数组，然后打印最大子数组的长度。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
 
// Function to return the maximum length
// of the sub-array such that the
// absolute difference between every two
// consecutive elements is either 1 or 0
int getMaxLength(int arr[],int n)
{
    int l = n;
    int i = 0, maxlen = 0;
    while (i < l)
    {
        int j = i;
        while (i+1 < l &&
             (abs(arr[i] - arr[i + 1]) == 1 ||
             abs(arr[i] - arr[i + 1]) == 0))
        {
            i++;
        }
 
            // Length of the valid sub-array currently
            // under consideration
            int currLen = i - j + 1;
 
            // Update the maximum length
            if (maxlen < currLen)
                maxlen = currLen;
 
            if (j == i)
                i++;
    }
 
    // Any valid sub-array cannot be of length 1
    //maxlen = (maxlen == 1) ? 0 : maxlen;
 
    // Return the maximum possible length
    return maxlen;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << getMaxLength(arr, n);
}
 
// This code is contributed by
// Surendra_Gangwar

Java

// Java implementation of the approach
public class GFG {
 
    // Function to return the maximum length
    // of the sub-array such that the
    // absolute difference between every two
    // consecutive elements is either 1 or 0
    public static int getMaxLength(int arr[])
    {
 
        int l = arr.length;
        int i = 0, maxlen = 0;
        while (i < l) {
            int j = i;
            while (i + 1 < l
                   && (Math.abs(arr[i] - arr[i + 1]) == 1
                       || Math.abs(arr[i] - arr[i + 1]) == 0)) {
                i++;
            }
 
            // Length of the valid sub-array currently
            // under cosideration
            int currLen = i - j + 1;
 
            // Update the maximum length
            if (maxlen < currLen)
                maxlen = currLen;
 
            if (j == i)
                i++;
        }
 
        // Any valid sub-array cannot be of length 1
        maxlen = (maxlen == 1) ? 0 : maxlen;
 
        // Return the maximum possible length
        return maxlen;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 4 };
        System.out.print(getMaxLength(arr));
    }
}

Python3

# Python3 implementation of the approach
 
# Function to return the maximum length
# of the sub-array such that the
# absolute difference between every two
# consecutive elements is either 1 or 0
def getMaxLength(arr, n) :
     
    l = n;
    i = 0; maxlen = 0;
     
    while (i < l) :
        j = i;
        while (i + 1 < l and
              (abs(arr[i] - arr[i + 1]) == 1 or
               abs(arr[i] - arr[i + 1]) == 0)) :
         
            i += 1;
         
        # Length of the valid sub-array
        # currently under cosideration
        currLen = i - j + 1;
 
        # Update the maximum length
        if (maxlen < currLen) :
            maxlen = currLen;
 
        if (j == i) :
            i += 1;
     
    # Any valid sub-array cannot be of length 1
    # maxlen = (maxlen == 1) ? 0 : maxlen;
 
    # Return the maximum possible length
    return maxlen;
     
# Driver code
if __name__ == "__main__" :
 
    arr = [ 2, 4 ];
    n = len(arr)
    print(getMaxLength(arr, n));
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the maximum length
    // of the sub-array such that the
    // Absolute difference between every two
    // consecutive elements is either 1 or 0
    public static int getMaxLength(int []arr)
    {
 
        int l = arr.Length;
        int i = 0, maxlen = 0;
        while (i < l)
        {
            int j = i;
            while (i + 1 < l &&
                    (Math.Abs(arr[i] - arr[i + 1]) == 1 ||
                    Math.Abs(arr[i] - arr[i + 1]) == 0))
            {
                i++;
            }
 
            // Length of the valid sub-array currently
            // under consideration
            int currLen = i - j + 1;
 
            // Update the maximum length
            if (maxlen < currLen)
                maxlen = currLen;
 
            if (j == i)
                i++;
        }
 
        // Any valid sub-array cannot be of length 1
        maxlen = (maxlen == 1) ? 0 : maxlen;
 
        // Return the maximum possible length
        return maxlen;
    }
 
    // Driver code
    public static void Main(String []args)
    {
        int []arr = { 2, 4 };
        Console.Write(getMaxLength(arr));
    }
}
 
// This code is contributed by Arnab Kundu

PHP

Javascript

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