最大总和连续子数组

编写一个有效的程序，以在具有最大和的一维数字数组中查找连续子数组的和。

Kadane的算法：

Initialize:
    max_so_far = INT_MIN
    max_ending_here = 0

Loop for each element of the array
  (a) max_ending_here = max_ending_here + a[i]
  (b) if(max_so_far < max_ending_here)
            max_so_far = max_ending_here
  (c) if(max_ending_here < 0)
            max_ending_here = 0
return max_so_far

解释：
Kadane算法的简单思想是查找数组的所有正连续段(为此使用max_ending_here)。并跟踪所有正分段中的最大和连续分段(为此使用max_so_far)。每次我们得到一个正和时，将其与max_so_far进行比较，如果max_so_far大于max_so_far，则对其进行更新

Lets take the example:
    {-2, -3, 4, -1, -2, 1, 5, -3}

    max_so_far = max_ending_here = 0

    for i=0,  a[0] =  -2
    max_ending_here = max_ending_here + (-2)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=1,  a[1] =  -3
    max_ending_here = max_ending_here + (-3)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=2,  a[2] =  4
    max_ending_here = max_ending_here + (4)
    max_ending_here = 4
    max_so_far is updated to 4 because max_ending_here greater 
    than max_so_far which was 0 till now

    for i=3,  a[3] =  -1
    max_ending_here = max_ending_here + (-1)
    max_ending_here = 3

    for i=4,  a[4] =  -2
    max_ending_here = max_ending_here + (-2)
    max_ending_here = 1

    for i=5,  a[5] =  1
    max_ending_here = max_ending_here + (1)
    max_ending_here = 2

    for i=6,  a[6] =  5
    max_ending_here = max_ending_here + (5)
    max_ending_here = 7
    max_so_far is updated to 7 because max_ending_here is 
    greater than max_so_far

    for i=7,  a[7] =  -3
    max_ending_here = max_ending_here + (-3)
    max_ending_here = 4

程序：

C++

// C++ program to print largest contiguous array sum
#include
#include
using namespace std;
 
int maxSubArraySum(int a[], int size)
{
    int max_so_far = INT_MIN, max_ending_here = 0;
 
    for (int i = 0; i < size; 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 maxSubArraySum*/
int main()
{
    int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
    int n = sizeof(a)/sizeof(a[0]);
    int max_sum = maxSubArraySum(a, n);
    cout << "Maximum contiguous sum is " << max_sum;
    return 0;
}

Java

import java.io.*;
// Java program to print largest contiguous array sum
import java.util.*;
 
class Kadane
{
    public static void main (String[] args)
    {
        int [] a = {-2, -3, 4, -1, -2, 1, 5, -3};
        System.out.println("Maximum contiguous sum is " +
                                       maxSubArraySum(a));
    }
 
    static int maxSubArraySum(int a[])
    {
        int size = a.length;
        int max_so_far = Integer.MIN_VALUE, max_ending_here = 0;
 
        for (int i = 0; i < size; 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;
    }
}

Python

# Python program to find maximum contiguous subarray
  
# Function to find the maximum contiguous subarray
from sys import maxint
def maxSubArraySum(a,size):
      
    max_so_far = -maxint - 1
    max_ending_here = 0
      
    for i in range(0, size):
        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 function to check the above function
a = [-13, -3, -25, -20, -3, -16, -23, -12, -5, -22, -15, -4, -7]
print "Maximum contiguous sum is", maxSubArraySum(a,len(a))
  
#This code is contributed by _Devesh Agrawal_

C#

// C# program to print largest
// contiguous array sum
using System;
 
class GFG
{
    static int maxSubArraySum(int []a)
    {
        int size = a.Length;
        int max_so_far = int.MinValue,
            max_ending_here = 0;
 
        for (int i = 0; i < size; 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 code
    public static void Main ()
    {
        int [] a = {-2, -3, 4, -1, -2, 1, 5, -3};
        Console.Write("Maximum contiguous sum is " +
                                maxSubArraySum(a));
    }
 
}
 
// This code is contributed by Sam007_

Javascript

 
// JavaScript program to find maximum
// contiguous subarray
  
// Function to find the maximum
// contiguous subarray
function maxSubArraySum(a, size)
{
    var maxint = Math.pow(2, 53)
    var max_so_far = -maxint - 1
    var max_ending_here = 0
      
    for (var i = 0; i < size; 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 code
var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ]
document.write("Maximum contiguous sum is",
               maxSubArraySum(a, a.length))
  
// This code is contributed by AnkThon
 

输出：

Maximum contiguous sum is 7

另一种方法：

C++

int maxSubarraySum(int arr[], int size)
{
    int max_ending_here = 0, max_so_far = INT_MIN;
    for (int i = 0; i < size; i++) {
       
        // include current element to previous subarray only
        // when it can add to a bigger number than itself.
        if (arr[i] <= max_ending_here + arr[i]) {
            max_ending_here += arr[i];
        }
       
        // Else start the max subarry from current element
        else {
            max_ending_here = arr[i];
        }
        if (max_ending_here > max_so_far)
            max_so_far = max_ending_here;
    }
    return max_so_far;
} // contributed by Vipul Raj

Java

static int maxSubArraySum(int a[],int size)
{
     
    int max_so_far = a[0], max_ending_here = 0;
 
    for (int i = 0; i < size; i++)
    {
        max_ending_here = max_ending_here + a[i];
        if (max_ending_here < 0)
            max_ending_here = 0;
         
        /* Do not compare for all
           elements. Compare only
           when max_ending_here > 0 */
        else if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
         
    }
    return max_so_far;
}
 
// This code is contributed by ANKITRAI1

Python

def maxSubArraySum(a,size):
     
    max_so_far = a[0]
    max_ending_here = 0
     
    for i in range(0, size):
        max_ending_here = max_ending_here + a[i]
        if max_ending_here < 0:
            max_ending_here = 0
         
        # Do not compare for all elements. Compare only  
        # when  max_ending_here > 0
        elif (max_so_far < max_ending_here):
            max_so_far = max_ending_here
             
    return max_so_far

C＃

static int maxSubArraySum(int[] a, int size)
{
    int max_so_far = a[0], max_ending_here = 0;
 
    for (int i = 0; i < size; i++) {
        max_ending_here = max_ending_here + a[i];
        if (max_ending_here < 0)
            max_ending_here = 0;
 
        /* Do not compare for all
        elements. Compare only
        when max_ending_here > 0 */
        else if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    return max_so_far;
}
 
// This code is contributed
// by ChitraNayal

的PHP

 0 */
    else if ($max_so_far < $max_ending_here)
        $max_so_far = $max_ending_here;
}
return $max_so_far;
 
// This code is contributed
// by ChitraNayal
?>

时间复杂度： O(n)
算法范例：动态编程
以下是Mohit Kumar建议的另一个简单实现。当数组中的所有数字均为负数时，该实现将处理这种情况。

C++

#include
using namespace std;
 
int maxSubArraySum(int a[], int size)
{
   int max_so_far = a[0];
   int curr_max = a[0];
 
   for (int i = 1; i < size; i++)
   {
        curr_max = max(a[i], curr_max+a[i]);
        max_so_far = max(max_so_far, curr_max);
   }
   return max_so_far;
}
 
/* Driver program to test maxSubArraySum */
int main()
{
   int a[] =  {-2, -3, 4, -1, -2, 1, 5, -3};
   int n = sizeof(a)/sizeof(a[0]);
   int max_sum = maxSubArraySum(a, n);
   cout << "Maximum contiguous sum is " << max_sum;
   return 0;
}

Java

// Java program to print largest contiguous
// array sum
import java.io.*;
 
class GFG {
 
    static int maxSubArraySum(int a[], int size)
    {
    int max_so_far = a[0];
    int curr_max = a[0];
 
    for (int i = 1; i < size; i++)
    {
           curr_max = Math.max(a[i], curr_max+a[i]);
        max_so_far = Math.max(max_so_far, curr_max);
    }
    return max_so_far;
    }
 
    /* Driver program to test maxSubArraySum */
    public static void main(String[] args)
    {
    int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
    int n = a.length;  
    int max_sum = maxSubArraySum(a, n);
    System.out.println("Maximum contiguous sum is "
                       + max_sum);
    }
}
 
// This code is contributd by Prerna Saini

Python

# Python program to find maximum contiguous subarray
 
def maxSubArraySum(a,size):
     
    max_so_far =a[0]
    curr_max = a[0]
     
    for i in range(1,size):
        curr_max = max(a[i], curr_max + a[i])
        max_so_far = max(max_so_far,curr_max)
         
    return max_so_far
 
# Driver function to check the above function
a = [-2, -3, 4, -1, -2, 1, 5, -3]
print"Maximum contiguous sum is" , maxSubArraySum(a,len(a))
 
#This code is contributed by _Devesh Agrawal_

C＃

// C# program to print largest
// contiguous array sum
using System;
 
class GFG
{
    static int maxSubArraySum(int []a, int size)
    {
    int max_so_far = a[0];
    int curr_max = a[0];
 
    for (int i = 1; i < size; i++)
    {
        curr_max = Math.Max(a[i], curr_max+a[i]);
        max_so_far = Math.Max(max_so_far, curr_max);
    }
 
    return max_so_far;
    }
 
    // Driver code
    public static void Main ()
    {
        int []a = {-2, -3, 4, -1, -2, 1, 5, -3};
        int n = a.Length;
        Console.Write("Maximum contiguous sum is "
                           + maxSubArraySum(a, n));
    }
 
}
 
// This code is contributed by Sam007_

输出：

Maximum contiguous sum is 7

为了打印具有最大和的子数组，只要获得最大和，我们就维护索引。

C++

// C++ program to print largest contiguous array sum
#include
#include
using namespace std;
 
int maxSubArraySum(int a[], int size)
{
    int max_so_far = INT_MIN, max_ending_here = 0,
       start =0, end = 0, s=0;
 
    for (int i=0; i< size; i++ )
    {
        max_ending_here += a[i];
 
        if (max_so_far < max_ending_here)
        {
            max_so_far = max_ending_here;
            start = s;
            end = i;
        }
 
        if (max_ending_here < 0)
        {
            max_ending_here = 0;
            s = i + 1;
        }
    }
    cout << "Maximum contiguous sum is "
        << max_so_far << endl;
    cout << "Starting index "<< start
        << endl << "Ending index "<< end << endl;
}
 
/*Driver program to test maxSubArraySum*/
int main()
{
    int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
    int n = sizeof(a)/sizeof(a[0]);
    int max_sum = maxSubArraySum(a, n);
    return 0;
}

Java

// Java program to print largest
// contiguous array sum
class GFG {
 
    static void maxSubArraySum(int a[], int size)
    {
        int max_so_far = Integer.MIN_VALUE,
        max_ending_here = 0,start = 0,
        end = 0, s = 0;
 
        for (int i = 0; i < size; i++)
        {
            max_ending_here += a[i];
 
            if (max_so_far < max_ending_here)
            {
                max_so_far = max_ending_here;
                start = s;
                end = i;
            }
 
            if (max_ending_here < 0)
            {
                max_ending_here = 0;
                s = i + 1;
            }
        }
        System.out.println("Maximum contiguous sum is "
                           + max_so_far);
        System.out.println("Starting index " + start);
        System.out.println("Ending index " + end);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { -2, -3, 4, -1, -2, 1, 5, -3 };
        int n = a.length;
        maxSubArraySum(a, n);
    }
}
 
// This code is contributed by  prerna saini

Python3

# Python program to print largest contiguous array sum
 
from sys import maxsize
 
# Function to find the maximum contiguous subarray
# and print its starting and end index
def maxSubArraySum(a,size):
 
    max_so_far = -maxsize - 1
    max_ending_here = 0
    start = 0
    end = 0
    s = 0
 
    for i in range(0,size):
 
        max_ending_here += a[i]
 
        if max_so_far < max_ending_here:
            max_so_far = max_ending_here
            start = s
            end = i
 
        if max_ending_here < 0:
            max_ending_here = 0
            s = i+1
 
    print ("Maximum contiguous sum is %d"%(max_so_far))
    print ("Starting Index %d"%(start))
    print ("Ending Index %d"%(end))
 
# Driver program to test maxSubArraySum
a = [-2, -3, 4, -1, -2, 1, 5, -3]
maxSubArraySum(a,len(a))

C＃

// C# program to print largest
// contiguous array sum
using System;
 
class GFG
{
    static void maxSubArraySum(int []a,
                               int size)
    {
        int max_so_far = int.MinValue,
        max_ending_here = 0, start = 0,
        end = 0, s = 0;
 
        for (int i = 0; i < size; i++)
        {
            max_ending_here += a[i];
 
            if (max_so_far < max_ending_here)
            {
                max_so_far = max_ending_here;
                start = s;
                end = i;
            }
 
            if (max_ending_here < 0)
            {
                max_ending_here = 0;
                s = i + 1;
            }
        }
        Console.WriteLine("Maximum contiguous " +
                         "sum is " + max_so_far);
        Console.WriteLine("Starting index " +
                                      start);
        Console.WriteLine("Ending index " +
                                      end);
    }
 
    // Driver code
    public static void Main()
    {
        int []a = {-2, -3, 4, -1,
                   -2, 1, 5, -3};
        int n = a.Length;
        maxSubArraySum(a, n);
    }
}
 
// This code is contributed
// by anuj_67.

输出：

Maximum contiguous sum is 7
Starting index 2
Ending index 6

时间复杂度： O(n)

辅助空间： O(1)

现在尝试下面的问题
给定一个整数数组(可能有些元素为负数)，编写一个C程序，通过将n个连续数组中的n个连续整数乘以n <= ARRAY_SIZE来找出最大乘积。同时打印最大乘积子数组的起点。