最大和连续子数组的Python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：

Python3

# Python program to find maximum contiguous subarray
  
# Function to find the maximum contiguous subarray
from math import inf
maxint=inf
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_

输出：

Maximum contiguous sum is 7

另一种方法：

Python3

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

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。

Python3

# 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_

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。

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))

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，找出*最大乘积*。另外，打印最大乘积子数组的起点。