題目(5kyu)：

The maximum sum subarray problem consists in finding the maximum sum of a contiguous >subsequence in an array or list of integers:

max_sequence([-2, 1, -3, 4, -1, 2, 1, -5, 4])
# should be 6: [4, -1, 2, 1]

Easy case is when the list is made up of only positive numbers and the maximum sum is >the sum of the whole array. If the list is made up of only negative numbers, return 0 >instead.

Empty list is considered to have zero greatest sum. Note that the empty list or array >is also a valid sublist/subarray.

解題思路

題目理解：給定一個元素皆為數字的列表arr，找到該列表所有連續組合中最大總和，並返還該總和。如果列表內皆為負數或為空列表，則返還0。
所有的連續子集合必存在起始索引位置i＆結束位置索引k，故需利用巢狀迴圈找出所有i&k匹配組合，取其sum(arr[i:k])值最大者。

def max_sequence(arr):
    #若為空集合則返還0
    if arr == []:
        return 0
    all_sum = []
    #第一層for迴圈遍歷所有起始索引i的可能性
    for i in range(len(arr)):
        #第二層for迴圈遍歷所有結束索引k的可能性
        for k in range(len(arr)):
            #若k<i則該子即視作[]故不影響計算，須留意list[a:b]僅包含到b-1，故需+1
            all_sum.append(sum(arr[i:k+1]))
    #若最大和小於0，代表arr內皆為負數故返還0
    if max(all_sum)<0:
        return 0
    return max(all_sum)