class Solution:
    def FindGreatestSumOfSubArray(self, array: List[int]) -> List[int]:
        # First: find the best value
        best = array[0]
        sum_cum = 0
        for i in range(len(array)):
            sum_cum = max(sum_cum + array[i], array[i])
            best = max(best, sum_cum)
        # Then, find the best solution
        sum_cum = 0
        start_idx, end_idx = 0, 0
        solutions = {}
        for i in range(len(array)):
            if sum_cum + array[i] >= array[i]:
                sum_cum = sum_cum + array[i]
                end_idx = i
                if sum_cum == best:
                    solutions[(start_idx, end_idx)] = end_idx - start_idx + 1
            else:
                sum_cum = array[i]
                start_idx, end_idx = i, i
                if len(solutions) == 0 and sum_cum == best:
                    solutions[(start_idx, end_idx)] = end_idx - start_idx + 1
        start_idx, end_idx = max(solutions, key=solutions.get)
        return array[start_idx:end_idx+1]

解题思路：

首先找到最优的值 best，参考 连续子数组的最大和

第二步开始寻找满足最优值的所有解，存储在字典 solutions 中（键：最优解第一个元素和最后一个元素的下标的元组 (start_idx, end_idx)；值：最优解的长度）

每当当前的 sum_cum 等于 best 时，就存储 (start_idx, end_idx)；

注意：由于要找长度最长的的最优解（且这个解唯一），所以如果字典 solutions 中已经非空，并且 start_idx 等于 end_idx，此时可以不存储这个长度为 1 的解