Algorithms

# 先求解第三种跨越中点的情况
def find_max_crossing_subarry(A, low, mid, high):
    # 求解最大的A[L, mid](mid 始终不变, L >= i)
    left_sum = -float('inf')
    suml = 0
    max_left = mid
    for i in range(mid):
        suml = suml + A[mid-i]
        if suml > left_sum:
            left_sum = suml
            max_left = mid - i

    #求最大的A[mid, R](mid 始终不变, j >= R >= mid) 组成
    right_sum = -float('inf')
    sumr = 0
    max_right = mid
    for j in range(mid+1,high-1):
        sumr = sumr + A[j]
        if sumr > right_sum:
            right_sum = sumr
            max_right = j

    return max_left, max_right, left_sum + right_sum

# 递归求解 a，b 两种情况的的 最大子数组的 candidate, 通过跟请况 c 的最大子数组 candidate 对比, return 最大子数组.
def find_max_subarry(A, low, high):
    if high == low:
        return low, high, A[low]
    else:
        mid = ((low + high) // 2)

        # 递归求解情况 a
        left_low, left_high, left_sum = find_max_subarry(A, low, mid)
        
        # 递归求解情况 b
        right_low, right_high, right_sum = find_max_subarry(A, mid+1, high)
        # 求解情况 c (跨中点的情况)
        cross_low, cross_high, cross_sum = find_max_crossing_subarry(A, low, mid, high)

        # 通过对比 3 种情况中最大子数组的 candidate, 返回 candidate 中的'最大者' 即为所求的 '最大子数组'.
        print('Candiate A :', A[left_low: left_high] )
        print('Candiate B :', A[right_low: right_high])
        print('Candiate C :', A[cross_low: cross_high])
        if left_sum >= right_sum and left_sum >= cross_sum:
            return left_low, left_high, left_sum
        elif right_sum >= left_sum and right_sum >= cross_sum:
            return right_low, right_high, right_sum
        else:
            return cross_low, cross_high, cross_sum


if __name__ == '__main__':

    A = [13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7]
  
    a,b,c = find_max_subarry(A, 0, 15)

    print('The maximum subarry is ', A[a:b+1])
    print('The aximum profit is %d . Buy in on day %d, and sale out on day %d .'% (c, a+1, b+1 ))

结果打印

Candiate A : []                    # 情况 a 的 最大子数组 candidate
Candiate C : [20]
Candiate C : [20, -3, -16]
Candiate C : [20, -3]
Candiate C : [20]
Candiate C : [18]
Candiate C : [18, 20, -7]
Candiate C : [18, 20]
Candiate C : [18, 20, -7, 12, -5]
Candiate C : [18, 20, -7, 12, -5, -22, 15]  # 情况 c 的 最大子数组 candidate
Candiate C : [18, 20, -7, 12, -5, -22]
Candiate C : [18, 20, -7, 12, -5, -22]
Candiate B : [18, 20, -7, 12, -5, -22]        # 情况 b 的 最大子数组 candidate
Candiate C : [18, 20, -7]

# 结果
The maximum subarry is  [18, 20, -7, 12]
The aximum profit is 43 . Buy in on day 8, and sale out on day 11 .

Reference

　　1，Introduction to algorithms