最大长度子序列，相邻元素之间的差异为 0 或 1

给定一个由n 个整数组成的数组。问题是找到相邻元素之间的差异为 0 或 1 的子序列的最大长度。
例子：

Input : arr[] = {2, 5, 6, 3, 7, 6, 5, 8}
Output : 5
The subsequence is {5, 6, 7, 6, 5}.

Input : arr[] = {-2, -1, 5, -1, 4, 0, 3}
Output : 4
The subsequence is {-2, -1, -1, 0}.

来源： Expedia 采访经历 | 12套

此问题的解决方案与最长递增子序列问题非常相似。唯一的区别是这里我们要检查子序列相邻元素之间的绝对差是0还是1。

C++

// C++ implementation to find maximum length
// subsequence with difference between adjacent
// elements as either 0 or 1
#include 
using namespace std;
 
// function to find maximum length subsequence
// with difference between adjacent elements as
// either 0 or 1
int maxLenSub(int arr[], int n)
{
    int mls[n], max = 0;
     
    // Initialize mls[] values for all indexes
    for (int i=0; iJava

// JAVA Code for Maximum length subsequence
// with difference between adjacent elements
// as either 0 or 1
import java.util.*;
 
class GFG {
     
    // function to find maximum length subsequence
    // with difference between adjacent elements as
    // either 0 or 1
     public static int maxLenSub(int arr[], int n)
    {
        int mls[] = new int[n], max = 0;
          
        // Initialize mls[] values for all indexes
        for (int i = 0; i < n; i++)
            mls[i] = 1;
          
        // Compute optimized maximum length
        // subsequence values in bottom up manner
        for (int i = 1; i < n; i++)
            for (int j = 0; j < i; j++)
                if (Math.abs(arr[i] - arr[j]) <= 1
                      && mls[i] < mls[j] + 1)
                    mls[i] = mls[j] + 1;
          
        // Store maximum of all 'mls' values in 'max'   
        for (int i = 0; i < n; i++)
            if (max < mls[i])
                max = mls[i];
          
        // required maximum length subsequence
        return max;       
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr[] = {2, 5, 6, 3, 7, 6, 5, 8};
        int n = arr.length;
        System.out.println("Maximum length subsequence = "+
                               maxLenSub(arr, n));
           
    }
}
   
// This code is contributed by Arnav Kr. Mandal.



Python3

# Python implementation to find maximum length
# subsequence with difference between adjacent
# elements as either 0 or 1
 
# function to find maximum length subsequence
# with difference between adjacent elements as
# either 0 or 1
def maxLenSub( arr, n):
    mls=[]
    max = 0
     
    #Initialize mls[] values for all indexes
    for i in range(n):
        mls.append(1)
     
    #Compute optimized maximum length subsequence
    # values in bottom up manner
    for i in range(n):
        for j in range(i):
            if (abs(arr[i] - arr[j]) <= 1 and mls[i] < mls[j] + 1):
                mls[i] = mls[j] + 1
                 
    # Store maximum of all 'mls' values in 'max'
    for i in range(n):
        if (max < mls[i]):
            max = mls[i]
     
    #required maximum length subsequence
    return max
     
#Driver program to test above
arr = [2, 5, 6, 3, 7, 6, 5, 8]
n = len(arr)
print("Maximum length subsequence = ",maxLenSub(arr, n))
 
#This code is contributed by "Abhishek Sharma 44"



C#

// C# Code for Maximum length subsequence
// with difference between adjacent elements
// as either 0 or 1
using System;
 
class GFG {
     
    // function to find maximum length subsequence
    // with difference between adjacent elements as
    // either 0 or 1
    public static int maxLenSub(int[] arr, int n)
    {
        int[] mls = new int[n];
        int max = 0;
 
        // Initialize mls[] values for all indexes
        for (int i = 0; i < n; i++)
            mls[i] = 1;
 
        // Compute optimized maximum length
        // subsequence values in bottom up manner
        for (int i = 1; i < n; i++)
            for (int j = 0; j < i; j++)
                if (Math.Abs(arr[i] - arr[j]) <= 1
                    && mls[i] < mls[j] + 1)
                    mls[i] = mls[j] + 1;
 
        // Store maximum of all 'mls' values in 'max'
        for (int i = 0; i < n; i++)
            if (max < mls[i])
                max = mls[i];
 
        // required maximum length subsequence
        return max;
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int[] arr = { 2, 5, 6, 3, 7, 6, 5, 8 };
        int n = arr.Length;
        Console.Write("Maximum length subsequence = " +
                                    maxLenSub(arr, n));
    }
}
 
// This code is contributed by Sam007







Javascript

 
// Javascript Code for
// Maximum length subsequence
// with difference between
// adjacent elements
// as either 0 or 1
 
    // function to find maximum
    // length subsequence
    // with difference between
    // adjacent elements as
    // either 0 or 1
     function  maxLenSub(arr, n)
    {
        let mls = new Array(n).fill(1), max = 0;
          
        // Initialize mls[] values for all indexes
        for (let i = 0; i < n; i++)
            mls[i] = 1;
          
        // Compute optimized maximum length
        // subsequence values in bottom up manner
        for (let i = 1; i < n; i++)
            for (let j = 0; j < i; j++)
                if (Math.abs(arr[i] - arr[j]) <= 1
                      && mls[i] < mls[j] + 1)
                    mls[i] = mls[j] + 1;
          
        // Store maximum of all 'mls' values in 'max'   
        for (let i = 0; i < n; i++)
            if (max < mls[i])
                max = mls[i];
          
        // required maximum length subsequence
        return max;       
    }
     
     
 
// driver program
 
        let arr = [2, 5, 6, 3, 7, 6, 5, 8];
        let n = arr.length;
        document.write("Maximum length subsequence = "+
                               maxLenSub(arr, n));
         

输出：

Maximum length subsequence = 5

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