圆形数组中的最大和，使得没有两个元素相邻

给定一个包含正整数值的圆形数组。任务是找到一个子序列的最大和，约束条件是序列中没有 2 个数字应该在数组中相邻。
例子：

Input: circular arr = {1, 2, 3, 1}
Output : 4
subsequence will be(1, 3), hence 1 + 3 = 4 

Input: circular arr = {1, 2, 3, 4, 5, 1}
Output: 9
subsequence will be(1, 3, 5), hence 1 + 3 + 5 = 9 

方法该问题可以使用 DP 解决。这篇文章中已经讨论了一种方法，但它用于数组。我们可以将循环子数组处理为两个数组，一个从 (0th 到 n-2-th) 和 (1st 到 n-1-th) 索引，并使用上一篇文章中使用的方法。两者返回的最大总和将是答案。

下面是上述方法的实现。

C++

// CPP program to find maximum sum in a circular array
// such that no elements are adjacent in the sum.
#include 
using namespace std;
 
// Function to calculate the sum
// from 0th position to(n-2)th position
int maxSum1(int arr[], int n)
{
    int dp[n];
    int maxi = 0;
 
    for (int i = 0; i < n - 1; i++) {
 
        // copy the element of original array to dp[]
        dp[i] = arr[i];
 
        // find the maximum element in the array
        if (maxi < arr[i])
            maxi = arr[i];
    }
 
    // start from 2nd to n-1th pos
    for (int i = 2; i < n - 1; i++) {
 
        // traverse for all pairs
        // bottom-up approach
        for (int j = 0; j < i - 1; j++) {
 
            // dp-condition
            if (dp[i] < dp[j] + arr[i]) {
                dp[i] = dp[j] + arr[i];
 
                // find maximum sum
                if (maxi < dp[i])
                    maxi = dp[i];
            }
        }
    }
 
    // return the maximum
    return maxi;
}
 
// Function to find the maximum sum
// from 1st position to n-1-th position
int maxSum2(int arr[], int n)
{
    int dp[n];
    int maxi = 0;
 
    for (int i = 1; i < n; i++) {
        dp[i] = arr[i];
 
        if (maxi < arr[i])
            maxi = arr[i];
    }
 
    // Traverse from third to n-th pos
    for (int i = 3; i < n; i++) {
 
        // bootom-up approach
        for (int j = 1; j < i - 1; j++) {
 
            // dp condition
            if (dp[i] < arr[i] + dp[j]) {
                dp[i] = arr[i] + dp[j];
 
                // find max sum
                if (maxi < dp[i])
                    maxi = dp[i];
            }
        }
    }
 
    // return max
    return maxi;
}
 
int findMaxSum(int arr[], int n)
{
   return max(maxSum1(arr, n), maxSum2(arr, n));
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 1 };
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << findMaxSum(arr, n);
    return 0;
}

Java

// Java  program to find maximum sum in a circular array
// such that no elements are adjacent in the sum.
 
import java.io.*;
 
class GFG {
     
     
// Function to calculate the sum
// from 0th position to(n-2)th position
static int maxSum1(int arr[], int n)
{
    int dp[]=new int[n];
    int maxi = 0;
 
    for (int i = 0; i < n - 1; i++) {
 
        // copy the element of original array to dp[]
        dp[i] = arr[i];
 
        // find the maximum element in the array
        if (maxi < arr[i])
            maxi = arr[i];
    }
 
    // start from 2nd to n-1th pos
    for (int i = 2; i < n - 1; i++) {
 
        // traverse for all pairs
        // bottom-up approach
        for (int j = 0; j < i - 1; j++) {
 
            // dp-condition
            if (dp[i] < dp[j] + arr[i]) {
                dp[i] = dp[j] + arr[i];
 
                // find maximum sum
                if (maxi < dp[i])
                    maxi = dp[i];
            }
        }
    }
 
    // return the maximum
    return maxi;
}
 
// Function to find the maximum sum
// from 1st position to n-1-th position
static int maxSum2(int arr[], int n)
{
    int dp[]=new int[n];
    int maxi = 0;
 
    for (int i = 1; i < n; i++) {
        dp[i] = arr[i];
 
        if (maxi < arr[i])
            maxi = arr[i];
    }
 
    // Traverse from third to n-th pos
    for (int i = 3; i < n; i++) {
 
        // bootom-up approach
        for (int j = 1; j < i - 1; j++) {
 
            // dp condition
            if (dp[i] < arr[i] + dp[j]) {
                dp[i] = arr[i] + dp[j];
 
                // find max sum
                if (maxi < dp[i])
                    maxi = dp[i];
            }
        }
    }
 
    // return max
    return maxi;
}
 
static int findMaxSum(int arr[], int n)
{
    int t=Math.max(maxSum1(arr, n), maxSum2(arr, n));
    return  t;
}
 
// Driver Code
    public static void main (String[] args) {
         
        int arr[] = { 1, 2, 3, 1 };
        int n = arr.length;
        System.out.println(findMaxSum(arr, n));
         
    }
}

Python 3

# Python 3 program to find maximum sum
# in a circular array such that no
# elements are adjacent in the sum.
 
# Function to calculate the sum from
# 0th position to(n-2)th position
def maxSum1(arr, n):
 
    dp = [0] * n
    maxi = 0
 
    for i in range(n - 1):
 
        # copy the element of original
        # array to dp[]
        dp[i] = arr[i]
 
        # find the maximum element in the array
        if (maxi < arr[i]):
            maxi = arr[i]
 
    # start from 2nd to n-1th pos
    for i in range(2, n - 1):
 
        # traverse for all pairs bottom-up
        # approach
        for j in range(i - 1) :
 
            # dp-condition
            if (dp[i] < dp[j] + arr[i]):
                dp[i] = dp[j] + arr[i]
 
                # find maximum sum
                if (maxi < dp[i]):
                    maxi = dp[i]
 
    # return the maximum
    return maxi
 
# Function to find the maximum sum
# from 1st position to n-1-th position
def maxSum2(arr, n):
 
    dp = [0] * n
    maxi = 0
 
    for i in range(1, n):
        dp[i] = arr[i]
 
        if (maxi < arr[i]):
            maxi = arr[i]
 
    # Traverse from third to n-th pos
    for i in range(3, n):
 
        # bootom-up approach
        for j in range(1, i - 1) :
 
            # dp condition
            if (dp[i] < arr[i] + dp[j]):
                dp[i] = arr[i] + dp[j]
 
                # find max sum
                if (maxi < dp[i]):
                    maxi = dp[i]
 
    # return max
    return maxi
 
def findMaxSum(arr, n):
    return max(maxSum1(arr, n), maxSum2(arr, n))
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 1, 2, 3, 1 ]
    n = len(arr)
    print(findMaxSum(arr, n))
 
# This code is contributed by ita_c

C#

// C# program to find maximum sum
// in a circular array such that
// no elements are adjacent in the sum.
using System;
 
class GFG
{
// Function to calculate the sum
// from 0th position to(n-2)th position
static int maxSum1(int []arr, int n)
{
    int []dp = new int[n];
    int maxi = 0;
 
    for (int i = 0; i < n - 1; i++)
    {
 
        // copy the element of original
        // array to dp[]
        dp[i] = arr[i];
 
        // find the maximum element
        // in the array
        if (maxi < arr[i])
            maxi = arr[i];
    }
 
    // start from 2nd to n-1th pos
    for (int i = 2; i < n - 1; i++)
    {
 
        // traverse for all pairs
        // bottom-up approach
        for (int j = 0; j < i - 1; j++)
        {
 
            // dp-condition
            if (dp[i] < dp[j] + arr[i])
            {
                dp[i] = dp[j] + arr[i];
 
                // find maximum sum
                if (maxi < dp[i])
                    maxi = dp[i];
            }
        }
    }
 
    // return the maximum
    return maxi;
}
 
// Function to find the maximum sum
// from 1st position to n-1-th position
static int maxSum2(int []arr, int n)
{
    int []dp = new int[n];
    int maxi = 0;
 
    for (int i = 1; i < n; i++)
    {
        dp[i] = arr[i];
 
        if (maxi < arr[i])
            maxi = arr[i];
    }
 
    // Traverse from third to n-th pos
    for (int i = 3; i < n; i++)
    {
 
        // bootom-up approach
        for (int j = 1; j < i - 1; j++)
        {
 
            // dp condition
            if (dp[i] < arr[i] + dp[j])
            {
                dp[i] = arr[i] + dp[j];
 
                // find max sum
                if (maxi < dp[i])
                    maxi = dp[i];
            }
        }
    }
 
    // return max
    return maxi;
}
 
static int findMaxSum(int []arr, int n)
{
    int t = Math.Max(maxSum1(arr, n),
                     maxSum2(arr, n));
    return t;
}
 
// Driver Code
static public void Main ()
{
    int []arr = { 1, 2, 3, 1 };
    int n = arr.Length;
    Console.WriteLine(findMaxSum(arr, n));
}
}
 
// This code is contributed
// by Sach_Code

Javascript

 
// JavaScript program to find maximum sum
// in a circular array such that
// no elements are adjacent in the sum.
 
// Function to calculate the sum
// from 0th position to(n-2)th position
function maxSum1(arr, n)
{
    let dp = new Array(n);
    let maxi = 0;
 
    for (i = 0; i < n - 1; i++)
    {
 
        // copy the element of original
        // array to dp[]
        dp[i] = arr[i];
 
        // find the maximum element
        // in the array
        if (maxi < arr[i])
            maxi = arr[i];
    }
 
    // start from 2nd to n-1th pos
    for (i = 2; i < n - 1; i++)
    {
 
        // traverse for all pairs
        // bottom-up approach
        for (j = 0; j < i - 1; j++)
        {
 
            // dp-condition
            if (dp[i] < dp[j] + arr[i])
            {
                dp[i] = dp[j] + arr[i];
 
                // find maximum sum
                if (maxi < dp[i])
                    maxi = dp[i];
            }
        }
    }
 
    // return the maximum
    return maxi;
}
 
// Function to find the maximum sum
// from 1st position to n-1-th position
function maxSum2(arr, n)
{
    let dp = new Array(n);
    let maxi = 0;
 
    for (i = 1; i < n; i++)
    {
        dp[i] = arr[i];
 
        if (maxi < arr[i])
            maxi = arr[i];
    }
 
    // Traverse from third to n-th pos
    for (i = 3; i < n; i++)
    {
 
        // bootom-up approach
        for (j = 1; j < i - 1; j++)
        {
 
            // dp condition
            if (dp[i] < arr[i] + dp[j])
            {
                dp[i] = arr[i] + dp[j];
 
                // find max sum
                if (maxi < dp[i])
                    maxi = dp[i];
            }
        }
    }
 
    // return max
    return maxi;
}
 
function findMaxSum(arr, n)
{
    let t = Math.max(maxSum1(arr, n),
                     maxSum2(arr, n));
    return t;
}
 
// Driver Code
 
let arr = [1, 2, 3, 1 ];
let n = arr.length;
document.write(findMaxSum(arr, n));
 
// This code is contributed by mohit kumar 29.

输出：

4

时间复杂度： O(N^2)

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