找到最长的子数组，使得相邻元素之间的差异为 K

给定一个大小为N的数组arr[]和整数K 。任务是找到相邻元素之间的差异为K的最长子数组。

例子：

Input: arr[] = { 5, 5, 5, 10, 8, 6, 12, 13 }, K =1
Output: {12, 13}
Explanation: This is the longest subarray with difference between adjacents as 1.

Input: arr[] = {4, 6, 8, 9, 8, 12, 14, 17, 15}, K = 2
Output: {4, 6, 8}

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，并存储它的长度和起点。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并继续更新最大长度和起点。重复该过程，直到找到所有有效的子数组，然后打印最大长度的子数组。

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
// sub-array such that the
// absolute difference between every two
// consecutive elements is K
void getMaxLengthSubarray(int arr[],
                          int N, int K)
{
    int l = N;
    int i = 0, maxlen = 0;
    int max_len_start, max_len_end;
    while (i < l) {
        int j = i;
        while (i + 1 < l
               && (abs(arr[i]
                       - arr[i + 1]) == K)) {
            i++;
        }
 
        // Length of the valid sub-array
        // currently under consideration
        int currLen = i - j + 1;
 
        // Update the maximum length subarray
        if (maxlen < currLen) {
            maxlen = currLen;
            max_len_start = j;
            max_len_end = i;
        }
 
        if (j == i)
            i++;
    }
 
    // Print the maximum length subarray
    for (int p = max_len_start;
         p <= max_len_end; p++)
        cout << arr[p] << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 4, 6, 8, 9, 8, 12,
                 14, 17, 15 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
    getMaxLengthSubarray(arr, N, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
 
// Function to return the maximum length
// sub-array such that the
// absolute difference between every two
// consecutive elements is K
static void getMaxLengthSubarray(int arr[],
                          int N, int K)
{
    int l = N;
    int i = 0, maxlen = 0;
    int max_len_start = 0, max_len_end = 0;
    while (i < l) {
        int j = i;
        while (i + 1 < l
               && (Math.abs(arr[i]
                       - arr[i + 1]) == K)) {
            i++;
        }
 
        // Length of the valid sub-array
        // currently under consideration
        int currLen = i - j + 1;
 
        // Update the maximum length subarray
        if (maxlen < currLen) {
            maxlen = currLen;
            max_len_start = j;
            max_len_end = i;
        }
 
        if (j == i)
            i++;
    }
 
    // Print the maximum length subarray
    for (int p = max_len_start;
         p <= max_len_end; p++)
        System.out.print(arr[p] + " ");
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 4, 6, 8, 9, 8, 12,
                 14, 17, 15 };
    int K = 2;
    int N =  arr.length; 
    getMaxLengthSubarray(arr, N, K);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program to implement
# the above approach
 
# Function to return the maximum length
# sub-array such that the
# absolute difference between every two
# consecutive elements is K
def getMaxLengthSubarray(arr, N, K) :
     
    l = N
    i = 0
    maxlen = 0
    while (i < l) :
        j = i
        while (i + 1 < l
               and (abs(arr[i]
                       - arr[i + 1]) == K)) :
            i += 1
         
        # Length of the valid sub-array
        # currently under consideration
        currLen = i - j + 1
 
        # Update the maximum length subarray
        if (maxlen < currLen) :
            maxlen = currLen
            max_len_start = j
            max_len_end = i
         
        if (j == i) :
            i += 1
     
    # Print the maximum length subarray
    for p in range(max_len_start, max_len_end+1, 1) :
        print(arr[p], end=" ")
 
# Driver code
arr = [ 4, 6, 8, 9, 8, 12,
                 14, 17, 15 ]
K = 2
N = len(arr)
getMaxLengthSubarray(arr, N, K)
 
# This code is contributed by avijitmondal1998

C#

// C# program for the above approach
using System;
class GFG
{
 
// Function to return the maximum length
// sub-array such that the
// absolute difference between every two
// consecutive elements is K
static void getMaxLengthSubarray(int []arr,
                          int N, int K)
{
    int l = N;
    int i = 0, maxlen = 0;
    int max_len_start = 0, max_len_end = 0;
    while (i < l) {
        int j = i;
        while (i + 1 < l
               && (Math.Abs(arr[i]
                       - arr[i + 1]) == K)) {
            i++;
        }
 
        // Length of the valid sub-array
        // currently under consideration
        int currLen = i - j + 1;
 
        // Update the maximum length subarray
        if (maxlen < currLen) {
            maxlen = currLen;
            max_len_start = j;
            max_len_end = i;
        }
 
        if (j == i)
            i++;
    }
 
    // Print the maximum length subarray
    for (int p = max_len_start;
         p <= max_len_end; p++)
        Console.Write(arr[p] + " ");
}
 
// Driver code
public static void Main()
{
    int []arr = { 4, 6, 8, 9, 8, 12,
                 14, 17, 15 };
    int K = 2;
    int N =  arr.Length; 
    getMaxLengthSubarray(arr, N, K);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

4 6 8

时间复杂度： O(N)
辅助空间： O(1)