列表中两个切片的最小最大值/最大值

Question

我有一个像[3,10,4,3,9,15,6,13]类的列表，我想找到两个不重叠的序列/序列，并通过取最大-最低值来获得最大值值。 它们必须是顺序的，因此您不能从1减去项目3。但是，可以从3减去索引1。

因此，在我的示例中，您将获得对[3,10]和[3,15]。 您如何以编程方式执行此操作。 这是我到目前为止所得到的。

蟒蛇：

l = [3,10,4,3,9,15,6,13]
pair1=[max(l), min(l)]
l.remove(max(l))
l.remove(min(l))
pair2=[max(l), min(l)]

哪一个当然不能满足我的要求，但是我不确定如何找到两对。 找到上面的代码对可以正常工作，但不能找到两个，因为通常会出现重叠的序列。

Answer 1

写这个很有趣：

import itertools as it

l = [3,10,4,3,9,15,6,13]

r=sorted(
    map(lambda x: (x[0:2], x[2:4]),
        sorted(it.imap(
            lambda x: (min(x[0:2]), max(x[0:2]), min(x[2:4]), max(x[2:4])),
            it.imap(lambda x: (l[x[0]], l[x[1]], l[x[2]], l[x[3]]),
                    it.combinations(range(len(l)), 4))),
               key=lambda x: -(x[1]-x[0]+x[3]-x[2])))[0],
    key=lambda x: x[0]-x[1])

print(r)

Answer 2

可能不是那么优雅，但是请尝试以下操作：

l = [3,10,4,3,9,15,6,13]

# list to store diffs
diff=[]
# calculate diffs
for i in range(0, len(l)-1):
    diff.append(l[i+1]-l[i])

# list to store diff sequences
results=[]

# Findinf the sequences
curr=0;
while curr<=(len(diff)-1):

    # list to store positive differences
    result=[]

    # llop while the difference is positive
    while diff[curr]>0 :

        # if it is a first element
        if len(result) == 0:
            # in 0th place store the diff sum
            result.append(diff[curr])
            # in 1st place store the sequence start index 
            result.append(curr)
            # in 2nd place store the sequence end index
            result.append(curr+1)
        else:
            # summ the sequnce diff
            result[0]+=diff[curr]
            # update sequence end index
            result[2]=curr+1
        # move to next element
        curr+=1
        # if reached end of array, break the loop 
        if curr>(len(diff)-1):
            break
    # if there was a sequence store it             
    if (len(result) > 0):
        results.append(result)
    else:
        # move to next element in case of there was no sequence
        curr+=1

# sort the results by sum in reverse order of the 0th element(sum) and normal order of 1st element(start index)
results.sort(key=lambda x: x[0]*10-x[1], reverse=True)
# print first 2 results
for i in range(0,2):
    print "[%d,%d]" % (l[results[i][1]],l[results[i][2]])

至少在您输入时，它会打印所需的结果：）