最大和连续子数组的Java程序

编写一个高效的程序，在具有最大和的一维数字数组中找到连续子数组的和。

Kadane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（max_ending_here 用于此）。并跟踪所有正段之间的最大和连续段（max_so_far 用于此）。每次我们得到一个正和时，将它与 max_so_far 进行比较，如果它大于 max_so_far 则更新 max_so_far

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：

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;
    }
}

输出：

Maximum contiguous sum is 7

另一种方法：

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

时间复杂度： O(n)

算法范式：动态规划
以下是Mohit Kumar建议的另一个简单实现。当数组中的所有数字都是负数时，该实现会处理这种情况。

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 contributed by Prerna Saini

输出：

Maximum contiguous sum is 7

为了打印具有最大和的子数组，我们在获得最大和时维护索引。

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

输出：

Maximum contiguous sum is 7
Starting index 2
Ending index 6

Kadane 的算法既可以被视为贪心算法，也可以被视为 DP。正如我们所看到的，我们正在保持整数的运行总和，当它小于 0 时，我们将其重置为 0（贪婪部分）。这是因为继续负总和比重新开始新范围更糟糕。现在它也可以被视为一个 DP，在每个阶段我们有 2 个选择：要么获取当前元素并继续之前的总和，要么重新开始一个新的范围。这两个选择都在实施过程中得到照顾。

时间复杂度： O(n)

辅助空间： O(1)

现在试试下面的问题
给定一个整数数组（可能某些元素为负数），编写一个 C 程序，通过将数组中 n ≤ ARRAY_SIZE 的“n”个连续整数相乘来找出*最大乘积*。另外，打印最大乘积子数组的起点。