Applications

A list of type integer prior nums = [1,2, -2, -1,5, -4], please find the product of the maximum of a contiguous subsequence in the sequence (the sequence comprises at least one number)

Solution 1

Ideas analysis

Analysis of the meaning of problems found, this question requirements: 1. The product of the maximum; 2. contiguous subsequence;
It is assumed that i is an index of successive sub initiation sequence; J consecutive sub-sequence ending index; for loop all possible traverse the contiguous subsequence, and calculates the product of the largest contiguous subsequence can;
about traversal process is:

When i is 0, j is desirable 0,1,2, ... len (nums) -1;
When i is 1, j preferably 1,2, ..., len (the nums) -1;
...
When i is len (nums) -1, j desirable len (nums) -1;

Note: if i == j indicates that it contains only a sequence number;

Code

from functools import reduce

num = [1,2,-2,-1,5,-4]
max = num[0]
max_index_start = 0
max_index_end = 0

for i in range(len(num)):
    for j in range(i, len(num)):
        if i == j:
            cur_num = num[i]
        else:
            cur_list = num[i:j+1]
            # 此处的recude函数表示先对集合中的第 1、2 个元素进行操作，
            #得到的结果再与下一个元素用function函数运算
            cur_num = reduce(lambda x,y:x*y,cur_list)
            
        if cur_num >= max:
            max = cur_num
            max_index_start = i
            max_index_end = j

print(max, max_index_start,max_index_end)

Execution results are as follows:

20 3 5

The above results indicated that the sequence of consecutive sub-sequences maximum product 20, consecutive sequence is nums [3] ~ nums [5]

Solution 2

Ideas analysis

Similarly, consecutive subframes is assumed that i is the subscript sequence starting; J consecutive sub-sequence ends subscript; there: MUL (i, J) MUL = (0, J) // MUL (0, i. 1- )
the above equation is interpreted: the product of a continuous subsequence = [StartElement: contiguous subsequence end element] // product [starting elements: a contiguous subsequence before starting element] the product of

EG: the nums = [1,2, -2, -1,5, -4] Solution 1 is clear from the results for the contiguous subsequence [-1.5, -4],
the above equation expands to: (-1) × 5 × (-4) = 1 × 2 × (-2) × (-1) × 5 × (-4) // 1 × 2 × (-2)

Analysis of the above formula, summarized as follows:

When mul (0, j) = 0 product, there is described a number 0, is restarted;
When mul (0, j) the product <0, in order to ensure maximum product, mul (0, i-1) should be negative (the same number is positive); In this case, should be looking for the maximum negative;
When mul (0, j) the product of> 0, in order to ensure maximum product, mul (0, i-1) should be as positive (positive number is the same); In this case, to find the minimum should be negative;

Code

def maxMul(nums):
    if not nums: return
    #起始变量
    #目前的累乘
    cur_mul = 1

    #前面最小的正数
    min_pos = 1

    #前面最大的负数
    max_neg = float("-inf")

    #结果
    result = float("-inf")

    for num in nums:
        cur_mul *= num

        if cur_mul > 0:
            result = max(result, cur_mul//min_pos)
            min_pos = min(min_pos, cur_mul)

        elif cur_mul < 0 :
            if max_neg != float("-inf"):
                result = max(result, cur_mul//max_neg)
            else:
                #如果是-inf,结果更新为当前值和结果的最大值
                result = max(result,num)
            #找最大的负数
            max_neg = max(max_neg, cur_mul)
        else:
            cur_mul = 1
            min_pos = 1
            max_neg = float("-inf")
            result = max(result,num)
    return result

data = [1,2,-2,0,5,-4]
print(maxMul(data))

data1 = [1,2,-2,-1,5,-4]
print(maxMul(data1))