最大圆形子数组和

给定 n 个数字（+ve 和 -ve），排列成一个圆圈，找到连续数字的最大和。

例子：

Input: a[] = {8, -8, 9, -9, 10, -11, 12}
Output: 22 (12 + 8 - 8 + 9 - 9 + 10)

Input: a[] = {10, -3, -4, 7, 6, 5, -4, -1} 
Output:  23 (7 + 6 + 5 - 4 -1 + 10) 

Input: a[] = {-1, 40, -14, 7, 6, 5, -4, -1}
Output: 52 (7 + 6 + 5 - 4 - 1 - 1 + 40)

方法一最大和可以有两种情况：

• 情况 1：对最大总和有贡献的元素的排列方式使得没有环绕。示例：{-10, 2, -1, 5}, {-2, 4, -1, 4, -1}。在这种情况下，Kadane 的算法将产生结果。
• 情况 2：对最大总和有贡献的元素被排列成有环绕。示例：{10, -12, 11}, {12, -5, 4, -8, 11}。在这种情况下，我们将包装更改为非包装。让我们看看如何。贡献元素的包装意味着非贡献元素的不包装，因此找出非贡献元素的总和并从总和中减去该总和。要找出非贡献的总和，请反转每个元素的符号，然后运行 Kadane 算法。
  我们的数组就像一个环，我们必须消除最大连续负数，这意味着倒置数组中的最大连续正数。最后，我们比较两种情况下得到的和，并返回两个和的最大值。

感谢ashishdey0提出这个解决方案。

以下是上述方法的实现。

C++

// C++ program for maximum contiguous circular sum problem
#include 
using namespace std;
 
// Standard Kadane's algorithm to
// find maximum subarray sum
int kadane(int a[], int n);
 
// The function returns maximum
// circular contiguous sum in a[]
int maxCircularSum(int a[], int n)
{
    // Case 1: get the maximum sum using standard kadane'
    // s algorithm
    int max_kadane = kadane(a, n);
     // if maximum sum using standard kadane' is less than 0
    if(max_kadane < 0)
      return max_kadane;
 
    // Case 2: Now find the maximum sum that includes
    // corner elements.
    int max_wrap = 0, i;
    for (i = 0; i < n; i++) {
        max_wrap += a[i]; // Calculate array-sum
        a[i] = -a[i]; // invert the array (change sign)
    }
 
    // max sum with corner elements will be:
    // array-sum - (-max subarray sum of inverted array)
    max_wrap = max_wrap + kadane(a, n);
 
    // The maximum circular sum will be maximum of two sums
    return (max_wrap > max_kadane) ? max_wrap : max_kadane;
}
 
// Standard Kadane's algorithm to find maximum subarray sum
// See https:// www.geeksforgeeks.org/archives/576 for details
int kadane(int a[], int n)
{
    int max_so_far = 0, max_ending_here = 0;
    int i;
    for (i = 0; i < n; i++) {
        max_ending_here = max_ending_here + a[i];
         
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
          if (max_ending_here < 0)
              max_ending_here = 0;
    }
    return max_so_far;
}
 
/* Driver program to test maxCircularSum() */
int main()
{
    int a[] = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << "Maximum circular sum is " << maxCircularSum(a, n) << endl;
    return 0;
}
 
// This is code is contributed by rathbhupendra

C

// C program for maximum contiguous circular sum problem
#include 
 
// Standard Kadane's algorithm to find maximum subarray
// sum
int kadane(int a[], int n);
 
// The function returns maximum circular contiguous sum
// in a[]
int maxCircularSum(int a[], int n)
{
    // Case 1: get the maximum sum using standard kadane'
    // s algorithm
    int max_kadane = kadane(a, n);
 
    // Case 2: Now find the maximum sum that includes
    // corner elements.
    int max_wrap = 0, i;
    for (i = 0; i < n; i++) {
        max_wrap += a[i]; // Calculate array-sum
        a[i] = -a[i]; // invert the array (change sign)
    }
 
    // max sum with corner elements will be:
    // array-sum - (-max subarray sum of inverted array)
    max_wrap = max_wrap + kadane(a, n);
 
    // The maximum circular sum will be maximum of two sums
    return (max_wrap > max_kadane) ? max_wrap : max_kadane;
}
 
// Standard Kadane's algorithm to find maximum subarray sum
// See https:// www.geeksforgeeks.org/archives/576 for details
int kadane(int a[], int n)
{
    int max_so_far = 0, max_ending_here = 0;
    int i;
    for (i = 0; i < n; i++) {
        max_ending_here = max_ending_here + a[i];
        if (max_ending_here < 0)
            max_ending_here = 0;
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    return max_so_far;
}
 
/* Driver program to test maxCircularSum() */
int main()
{
    int a[] = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
    int n = sizeof(a) / sizeof(a[0]);
    printf("Maximum circular sum is %dn",
           maxCircularSum(a, n));
    return 0;
}

Java

// Java program for maximum contiguous circular sum problem
import java.io.*;
import java.util.*;
 
class Solution{
    public static int kadane(int a[],int n){
        int res = 0;
        int x =  a[0];
        for(int i = 0; i < n; i++){
            res = Math.max(a[i],res+a[i]);
            x= Math.max(x,res);
        }
        return x;
    }
  //lets write a function for calculating max sum in circular manner as discuss above
    public static int reverseKadane(int a[],int n){
        int total = 0;
      //taking the total sum of the array elements
        for(int i = 0; i< n; i++){
            total +=a[i];
             
        }
      // inverting the array
        for(int i = 0; i

---

Python

# Python program for maximum contiguous circular sum problem
 
# Standard Kadane's algorithm to find maximum subarray sum
def kadane(a):
    Max = a[0]
    temp = Max
    for i in range(1,len(a)):
        temp += a[i]
        if temp < a[i]:
            temp = a[i]
        Max = max(Max,temp)
    return Max
 
# The function returns maximum circular contiguous sum in
# a[]
def maxCircularSum(a):
 
    n = len(a)
 
    # Case 1: get the maximum sum using standard kadane's
    # algorithm
    max_kadane = kadane(a)
 
    # Case 2: Now find the maximum sum that includes corner
    # elements.
    # You can do so by finding the maximum negative contiguous
    # sum
    # convert a to -ve 'a' and run kadane's algo
    neg_a = [-1*x for x in a]
    max_neg_kadane = kadane(neg_a)
 
    # Max sum with corner elements will be:
    # array-sum - (-max subarray sum of inverted array)
    max_wrap = -(sum(neg_a)-max_neg_kadane)
 
    # The maximum circular sum will be maximum of two sums
    res = max(max_wrap,max_kadane)
    return res if res != 0  else max_kadane
 
# Driver function to test above function
a = [11, 10, -20, 5, -3, -5, 8, -13, 10]
print "Maximum circular sum is", maxCircularSum(a)
 
# This code is contributed by Devesh Agrawal

C#

// C# program for maximum contiguous
// circular sum problem
using System;
 
class MaxCircularSum {
 
    // The function returns maximum circular
    // contiguous sum in a[]
    static int maxCircularSum(int[] a)
    {
        int n = a.Length;
 
        // Case 1: get the maximum sum using standard kadane'
        // s algorithm
        int max_kadane = kadane(a);
 
        // Case 2: Now find the maximum sum that includes
        // corner elements.
        int max_wrap = 0;
        for (int i = 0; i < n; i++) {
            max_wrap += a[i]; // Calculate array-sum
            a[i] = -a[i]; // invert the array (change sign)
        }
 
        // max sum with corner elements will be:
        // array-sum - (-max subarray sum of inverted array)
        max_wrap = max_wrap + kadane(a);
 
        // The maximum circular sum will be maximum of two sums
        return (max_wrap > max_kadane) ? max_wrap : max_kadane;
    }
 
    // Standard Kadane's algorithm to find maximum subarray sum
    // See https:// www.geeksforgeeks.org/archives/576 for details
    static int kadane(int[] a)
    {
        int n = a.Length;
        int max_so_far = 0, max_ending_here = 0;
        for (int i = 0; i < n; i++) {
            max_ending_here = max_ending_here + a[i];
            if (max_ending_here < 0)
                max_ending_here = 0;
            if (max_so_far < max_ending_here)
                max_so_far = max_ending_here;
        }
        return max_so_far;
    }
 
    // Driver code
    public static void Main()
    {
        int[] a = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
 
        Console.Write("Maximum circular sum is " + maxCircularSum(a));
    }
}
 
/* This code is contributed by vt_m*/

PHP

 $max_kadane)? $max_wrap: $max_kadane;
}
 
// Standard Kadane's algorithm to
// find maximum subarray sum
// See https://www.geeksforgeeks.org/archives/576 for details
function kadane($a, $n)
{
    $max_so_far = 0;
    $max_ending_here = 0;
    for ($i = 0; $i < $n; $i++)
    {
        $max_ending_here = $max_ending_here +$a[$i];
        if ($max_ending_here < 0)
            $max_ending_here = 0;
        if ($max_so_far < $max_ending_here)
            $max_so_far = $max_ending_here;
    }
    return $max_so_far;
}
 
    /* Driver code */
    $a = array(11, 10, -20, 5, -3, -5, 8, -13, 10);
    $n = count($a);
    echo "Maximum circular sum is ". maxCircularSum($a, $n);
 
// This code is contributed by rathbhupendra
?>

Javascript

    // javascript program for maximum contiguous circular sum problem
 
    function kadane(a , n) {
        var res = 0;
        var x = a[0];
        for (i = 0; i < n; i++) {
            res = Math.max(a[i], res + a[i]);
            x = Math.max(x, res);
        }
        return x;
    }
 
    // lets write a function for calculating max sum in circular manner as discuss
    // above
    function reverseKadane(a , n) {
        var total = 0;
        // taking the total sum of the array elements
        for (i = 0; i < n; i++) {
            total += a[i];
        }
        // inverting the array
        for (i = 0; i < n; i++) {
            a[i] = -a[i];
        }
        // finding min sum subarray
        var k = kadane(a, n);
        // max circular sum
        var ress = total + k;
        // to handle the case in which all elements are negative
        if (total == -k) {
            return total;
        }
        else {
            return ress;
        }
 
    }
 
     
    var a = [11, 10, -20, 5, -3, -5, 8, -13, 10];
    var n = 9;
    if (n == 1) {
        document.write("Maximum circular sum is " + a[0]);
    }
      else {
 
        document.write("Maximum circular sum is "
            + Math.max(kadane(a, n), reverseKadane(a, n)));
    }
 
// This code is contributed by todaysgaurav

输出：

Maximum circular sum is 31

复杂性分析：

• 时间复杂度： O(n)，其中 n 是输入数组中的元素数。
因为只需要对数组进行线性遍历。
• 辅助空间： O(1)。
因为不需要额外的空间。

请注意，如果所有数字都是负数，例如 {-1, -2, -3}，则上述算法不起作用。在这种情况下它返回 0。这种情况可以通过在运行上述算法之前添加预检查以查看所有数字是否为负来处理。

方法二
方法：在此方法中，修改 Kadane 的算法，找到一个最小的连续子数组和和最大的连续子数组和，然后检查 max_value 和总和减去 min_value 后剩下的值之间的最大值。
算法

1. 我们将计算给定数组的总和。
2. 我们将变量 curr_max、max_so_far、curr_min、min_so_far 声明为数组的第一个值。
3. 现在我们将使用 Kadane 算法来找到最大子数组和和最小子数组和。
4. 检查数组中的所有值：-
   1. 如果 min_so_far 等于 sum，即所有值都是负数，那么我们返回 max_so_far。
   2. 否则，我们将计算 max_so_far 和 (sum – min_so_far) 的最大值并返回。

下面给出上述方法的实现。

C++

// C++ program for maximum contiguous circular sum problem
#include 
using namespace std;
 
// The function returns maximum
// circular contiguous sum in a[]
int maxCircularSum(int a[], int n)
{
    // Corner Case
    if (n == 1)
        return a[0];
 
    // Initialize sum variable which store total sum of the array.
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += a[i];
    }
 
    // Initialize every variable with first value of array.
    int curr_max = a[0], max_so_far = a[0], curr_min = a[0], min_so_far = a[0];
 
    // Concept of Kadane's Algorithm
    for (int i = 1; i < n; i++) {
        // Kadane's Algorithm to find Maximum subarray sum.
        curr_max = max(curr_max + a[i], a[i]);
        max_so_far = max(max_so_far, curr_max);
 
        // Kadane's Algorithm to find Minimum subarray sum.
        curr_min = min(curr_min + a[i], a[i]);
        min_so_far = min(min_so_far, curr_min);
    }
 
    if (min_so_far == sum)
        return max_so_far;
 
    // returning the maximum value
    return max(max_so_far, sum - min_so_far);
}
 
/* Driver program to test maxCircularSum() */
int main()
{
    int a[] = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << "Maximum circular sum is " << maxCircularSum(a, n) << endl;
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
 
class GFG {
  public static int maxCircularSum(int a[], int n)
  {
    // Corner Case
    if (n == 1)
      return a[0];
 
    // Initialize sum variable which store total sum of
    // the array.
    int sum = 0;
    for (int i = 0; i < n; i++) {
      sum += a[i];
    }
 
    // Initialize every variable with first value of
    // array.
    int curr_max = a[0], max_so_far = a[0],
    curr_min = a[0], min_so_far = a[0];
 
    // Concept of Kadane's Algorithm
    for (int i = 1; i < n; i++)
    {
 
      // Kadane's Algorithm to find Maximum subarray
      // sum.
      curr_max = Math.max(curr_max + a[i], a[i]);
      max_so_far = Math.max(max_so_far, curr_max);
 
      // Kadane's Algorithm to find Minimum subarray
      // sum.
      curr_min = Math.min(curr_min + a[i], a[i]);
      min_so_far = Math.min(min_so_far, curr_min);
    }
    if (min_so_far == sum) {
      return max_so_far;
    }
 
    // returning the maximum value
    return Math.max(max_so_far, sum - min_so_far);
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int a[] = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
    int n = 9;
    System.out.println("Maximum circular sum is "
                       + maxCircularSum(a, n));
  }
}
 
// This code is contributed by aditya7409

Python3

# Python program for maximum contiguous circular sum problem
 
# The function returns maximum
# circular contiguous sum in a[]
def maxCircularSum(a, n):
     
    # Corner Case
    if (n == 1):
        return a[0]
 
    # Initialize sum variable which
    # store total sum of the array.
    sum = 0
    for i in range(n):
        sum += a[i]
 
    # Initialize every variable
    # with first value of array.
    curr_max = a[0]
    max_so_far = a[0]
    curr_min = a[0]
    min_so_far = a[0]
 
    # Concept of Kadane's Algorithm
    for i in range(1, n):
       
        # Kadane's Algorithm to find Maximum subarray sum.
        curr_max = max(curr_max + a[i], a[i])
        max_so_far = max(max_so_far, curr_max)
 
        # Kadane's Algorithm to find Minimum subarray sum.
        curr_min = min(curr_min + a[i], a[i])
        min_so_far = min(min_so_far, curr_min)
    if (min_so_far == sum):
        return max_so_far
 
    # returning the maximum value
    return max(max_so_far, sum - min_so_far)
 
# Driver code
a = [11, 10, -20, 5, -3, -5, 8, -13, 10]
n = len(a)
print("Maximum circular sum is", maxCircularSum(a, n))
 
# This code is contributes by subhammahato348

C＃

// C# program for maximum contiguous circular sum problem
using System;
class GFG
{
    public static int maxCircularSum(int[] a, int n)
    {
        // Corner Case
        if (n == 1)
            return a[0];
 
        // Initialize sum variable which store total sum of
        // the array.
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += a[i];
        }
 
        // Initialize every variable with first value of
        // array.
        int curr_max = a[0], max_so_far = a[0],
            curr_min = a[0], min_so_far = a[0];
 
        // Concept of Kadane's Algorithm
        for (int i = 1; i < n; i++)
        {
 
            // Kadane's Algorithm to find Maximum subarray
            // sum.
            curr_max = Math.Max(curr_max + a[i], a[i]);
            max_so_far = Math.Max(max_so_far, curr_max);
 
            // Kadane's Algorithm to find Minimum subarray
            // sum.
            curr_min = Math.Min(curr_min + a[i], a[i]);
            min_so_far = Math.Min(min_so_far, curr_min);
        }
        if (min_so_far == sum)
        {
            return max_so_far;
        }
 
        // returning the maximum value
        return Math.Max(max_so_far, sum - min_so_far);
    }
 
    // Driver code
    public static void Main()
    {
        int[] a = { 11, 10, -20, 5, -3, -5, 8, -13, 10 };
        int n = 9;
        Console.WriteLine("Maximum circular sum is "
                          + maxCircularSum(a, n));
    }
}
 
// This code is contributed by subhammahato348

Javascript

       // JavaScript program for the above approach
 
 
       // The function returns maximum
       // circular contiguous sum in a[]
       function maxCircularSum(a, n) {
           // Corner Case
           if (n == 1)
               return a[0];
 
           // Initialize sum variable which store total sum of the array.
           let sum = 0;
           for (let i = 0; i < n; i++) {
               sum += a[i];
           }
 
           // Initialize every variable with first value of array.
           let curr_max = a[0], max_so_far = a[0], curr_min = a[0], min_so_far = a[0];
 
           // Concept of Kadane's Algorithm
           for (let i = 1; i < n; i++) {
               // Kadane's Algorithm to find Maximum subarray sum.
               curr_max = Math.max(curr_max + a[i], a[i]);
               max_so_far = Math.max(max_so_far, curr_max);
 
               // Kadane's Algorithm to find Minimum subarray sum.
               curr_min = Math.min(curr_min + a[i], a[i]);
               min_so_far = Math.min(min_so_far, curr_min);
           }
 
           if (min_so_far == sum)
               return max_so_far;
 
           // returning the maximum value
           return Math.max(max_so_far, sum - min_so_far);
       }
 
       // Driver program to test maxCircularSum()
 
       let a = [11, 10, -20, 5, -3, -5, 8, -13, 10];
       let n = a.length;
       document.write("Maximum circular sum is " + maxCircularSum(a, n));
 
   // This code is contributed by Potta Lokesh
 
   

输出：

Maximum circular sum is 31

复杂性分析：

• 时间复杂度： O(n)，其中 n 是输入数组中的元素数。
因为只需要对数组进行线性遍历。
• 辅助空间： O(1)。
因为不需要额外的空间。