使得两个元素都不相邻的最大和

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。所以3 2 7 10应该返回13(3和10之和)，或者3 2 5 10 7应该返回15(3、5和7之和)。以最有效的方式回答问题。

例子 ：

Input : arr[] = {5, 5, 10, 100, 10, 5}
Output : 110

Input : arr[] = {1, 2, 3}
Output : 4

Input : arr[] = {1, 20, 3}
Output : 20

算法：
循环访问arr []中的所有元素，并维护两个和incl和excl，其中incl =包括前一个元素的最大和，而excl =除去前一个元素的最大和。

排除当前元素的最大和为max(incl，excl)，包括当前元素的最大和为excl + current元素(请注意，仅考虑excl是因为元素不能相邻)。

在循环结束时，返回最大值incl和excl。

例子：

arr[] = {5,  5, 10, 40, 50, 35}

  incl = 5 
  excl = 0

  For i = 1 (current element is 5)
  incl =  (excl + arr[i])  = 5
  excl =  max(5, 0) = 5

  For i = 2 (current element is 10)
  incl =  (excl + arr[i]) = 15
  excl =  max(5, 5) = 5

  For i = 3 (current element is 40)
  incl = (excl + arr[i]) = 45
  excl = max(5, 15) = 15

  For i = 4 (current element is 50)
  incl = (excl + arr[i]) = 65
  excl =  max(45, 15) = 45

  For i = 5 (current element is 35)
  incl =  (excl + arr[i]) = 80
  excl =  max(65, 45) = 65

And 35 is the last element. So, answer is max(incl, excl) =  80

感谢Debanjan提供代码。

执行：

C/C++

#include
  
/*Function to return max sum such that no two elements
 are adjacent */
int FindMaxSum(int arr[], int n)
{
  int incl = arr[0];
  int excl = 0;
  int excl_new;
  int i;
  
  for (i = 1; i < n; i++)
  {
     /* current max excluding i */
     excl_new = (incl > excl)? incl: excl;
  
     /* current max including i */
     incl = excl + arr[i];
     excl = excl_new;
  }
  
   /* return max of incl and excl */
   return ((incl > excl)? incl : excl);
}
  
/* Driver program to test above function */
int main()
{
  int arr[] = {5, 5, 10, 100, 10, 5};
  int n = sizeof(arr) / sizeof(arr[0]);
  printf("%d n", FindMaxSum(arr, n));
  return 0;
}

Java

class MaximumSum
{
    /*Function to return max sum such that no two elements
      are adjacent */
    int FindMaxSum(int arr[], int n)
    {
        int incl = arr[0];
        int excl = 0;
        int excl_new;
        int i;
  
        for (i = 1; i < n; i++)
        {
            /* current max excluding i */
            excl_new = (incl > excl) ? incl : excl;
  
            /* current max including i */
            incl = excl + arr[i];
            excl = excl_new;
        }
  
        /* return max of incl and excl */
        return ((incl > excl) ? incl : excl);
    }
  
    // Driver program to test above functions
    public static void main(String[] args)
    {
        MaximumSum sum = new MaximumSum();
        int arr[] = new int[]{5, 5, 10, 100, 10, 5};
        System.out.println(sum.FindMaxSum(arr, arr.length));
    }
}
  
// This code has been contributed by Mayank Jaiswal

Python

# Function to return max sum such that 
# no two elements are adjacent
def find_max_sum(arr):
    incl = 0
    excl = 0
     
    for i in arr:
          
        # Current max excluding i (No ternary in 
        # Python)
        new_excl = excl if excl>incl else incl
         
        # Current max including i
        incl = excl + i
        excl = new_excl
      
    # return max of incl and excl
    return (excl if excl>incl else incl)
  
# Driver program to test above function
arr = [5, 5, 10, 100, 10, 5]
print find_max_sum(arr)
  
# This code is contributed by Kalai Selvan

C#

/* Program to return max sum such that no 
two elements are adjacent */
using System;
  
class GFG {
      
    /* Function to return max sum such
    that no two elements are adjacent */
    static int FindMaxSum(int []arr, int n)
    {
        int incl = arr[0];
        int excl = 0;
        int excl_new;
        int i;
  
        for (i = 1; i < n; i++)
        {
            /* current max excluding i */
            excl_new = (incl > excl) ? 
                            incl : excl;
  
            /* current max including i */
            incl = excl + arr[i];
            excl = excl_new;
        }
  
        /* return max of incl and excl */
        return ((incl > excl) ? 
                            incl : excl);
    }
  
    // Driver program to test above 
    // functions
    public static void Main()
    {
        int []arr = new int[]{5, 5, 10,
                              100, 10, 5};
                                
        Console.Write(
             FindMaxSum(arr, arr.Length));
    }
}
  
// This code has been contributed by
// nitin mittal

PHP

 $excl)? $incl: $excl;
  
    // current max including i 
    $incl = $excl + $arr[$i];
    $excl = $excl_new;
}
  
// return max of incl and excl 
return (($incl > $excl)? $incl : $excl);
}
  
// Driver Code
$arr = array(5, 5, 10, 100, 10, 5);
$n = sizeof($arr);
echo FindMaxSum($arr, $n);
      
// This code is contributed by Ajit
?>

输出：

110

时间复杂度： O(n)

有关更多说明，请参阅从房屋中查找最大可能的盗窃价值。

现在，也对带有负数的数组尝试同样的问题。