给定一个正数数组,请找到一个子序列的最大和,并约束该序列中的任何2个数字都不应与该数组相邻。因此3 2 7 10应该返回13(3和10的总和),或者3 2 5 10 7应该返回15(3、5和7的总和)。
例子:
Input : arr[] = {3, 5, 3}
Output : 6
Explanation :
Selecting indexes 0 and 2 will maximise the sum
i.e 3+3 = 6
Input : arr[] = {2, 5, 2}
Output : 5
在上一篇文章中,我们已经讨论了解决此问题的有效方法。
但是,我们也可以使用动态编程方法解决此问题。
动态编程方法:让我们确定“ dp”的状态。令dp [i]是从索引“ i”开始并以索引“ N-1”结束的子数组的最大可能和。现在,我们必须找到此状态与低阶状态之间的递归关系。
在这种情况下,对于索引“ i”,我们将有两种选择。
1) Choose the current index:
In this case, the relation will be dp[i] = arr[i] + dp[i+2]
2) Skip the current index:
Relation will be dp[i] = dp[i+1]
我们将选择使我们的结果最大化的路径。
因此,最终关系将是:
dp[i] = max(dp[i+2]+arr[i], dp[i+1])
下面是上述方法的实现:
C++
// C++ program to implement above approach
#include
#define maxLen 10
using namespace std;
// variable to store states of dp
int dp[maxLen];
// variable to check if a given state
// has been solved
bool v[maxLen];
// Function to find the maximum sum subsequence
// such that no two elements are adjacent
int maxSum(int arr[], int i, int n)
{
// Base case
if (i >= n)
return 0;
// To check if a state has
// been solved
if (v[i])
return dp[i];
v[i] = 1;
// Required recurrence relation
dp[i] = max(maxSum(arr, i + 1, n),
arr[i] + maxSum(arr, i + 2, n));
// Returning the value
return dp[i];
}
// Driver code
int main()
{
int arr[] = { 12, 9, 7, 33 };
int n = sizeof(arr) / sizeof(int);
cout << maxSum(arr, 0, n);
return 0;
}
Java
// Java program to implement above approach
class GFG
{
static int maxLen = 10;
// variable to store states of dp
static int dp[] = new int[maxLen];
// variable to check if a given state
// has been solved
static boolean v[] = new boolean[maxLen];
// Function to find the maximum sum subsequence
// such that no two elements are adjacent
static int maxSum(int arr[], int i, int n)
{
// Base case
if (i >= n)
return 0;
// To check if a state has
// been solved
if (v[i])
return dp[i];
v[i] = true;
// Required recurrence relation
dp[i] = Math.max(maxSum(arr, i + 1, n),
arr[i] + maxSum(arr, i + 2, n));
// Returning the value
return dp[i];
}
// Driver code
public static void main(String args[])
{
int arr[] = { 12, 9, 7, 33 };
int n = arr.length;
System.out.println( maxSum(arr, 0, n));
}
}
// This code is contributed by Arnab Kundu
Python3
# Python 3 program to implement above approach
maxLen = 10
# variable to store states of dp
dp = [0 for i in range(maxLen)]
# variable to check if a given state
# has been solved
v = [0 for i in range(maxLen)]
# Function to find the maximum sum subsequence
# such that no two elements are adjacent
def maxSum(arr, i, n):
# Base case
if (i >= n):
return 0
# To check if a state has
# been solved
if (v[i]):
return dp[i]
v[i] = 1
# Required recurrence relation
dp[i] = max(maxSum(arr, i + 1, n),
arr[i] + maxSum(arr, i + 2, n))
# Returning the value
return dp[i]
# Driver code
if __name__ == '__main__':
arr = [12, 9, 7, 33]
n = len(arr)
print(maxSum(arr, 0, n))
# This code is contributed by
# Surendra_Gangwar
C#
// C# program to implement above approach
using System;
class GFG
{
static int maxLen = 10;
// variable to store states of dp
static int[] dp = new int[maxLen];
// variable to check if a given state
// has been solved
static bool[] v = new bool[maxLen];
// Function to find the maximum sum subsequence
// such that no two elements are adjacent
static int maxSum(int[] arr, int i, int n)
{
// Base case
if (i >= n)
return 0;
// To check if a state has
// been solved
if (v[i])
return dp[i];
v[i] = true;
// Required recurrence relation
dp[i] = Math.Max(maxSum(arr, i + 1, n),
arr[i] + maxSum(arr, i + 2, n));
// Returning the value
return dp[i];
}
// Driver code
public static void Main()
{
int[] arr = { 12, 9, 7, 33 };
int n = arr.Length;
Console.Write( maxSum(arr, 0, n));
}
}
// This code is contributed by ChitraNayal
PHP
= $n)
return 0;
// To check if a state has
// been solved
if ($GLOBALS['v'][$i])
return $GLOBALS['dp'][$i];
$GLOBALS['v'][$i] = 1;
// Required recurrence relation
$GLOBALS['dp'][$i] = max(maxSum($arr, $i + 1, $n),
$arr[$i] + maxSum($arr, $i + 2, $n));
// Returning the value
return $GLOBALS['dp'][$i];
}
// Driver code
$arr = array( 12, 9, 7, 33 );
$n = count($arr);
echo maxSum($arr, 0, $n);
// This code is contributed by AnkitRai01
?>
输出:
45