题目：
给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。进阶: 如果你已经实现复杂度为 O(n) 的解法，尝试使用更为精妙的分治法求解。
思路：贪心法和分而治之
程序1：贪心发
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
length = len(nums)
auxiliary = nums[0]
temp_result = nums[0]
for index in range(1,length):
auxiliary = max(nums[index], auxiliary + nums[index])
temp_result = max(auxiliary, temp_result)
result = temp_result
return result
程序2：分而治之
官方思路示图：
class Solution:
def maxSubArray(self, nums: List[int]) -> int:
length = len(nums)
if length <= 1:
return nums[0]
left = self.maxSubArray(nums[0 : len(nums) // 2])
right = self.maxSubArray(nums[len(nums) //2 : len(nums)])
middle_left = nums[len(nums) // 2 - 1]
auxiliary_1 = 0
for index1 in range(len(nums) // 2 - 1, -1, -1):
auxiliary_1 += nums[index1]
middle_left = max(middle_left, auxiliary_1)
middle_right = nums[len(nums) // 2]
auxiliary_2 = 0
for index2 in range(len(nums) // 2, len(nums), 1):
auxiliary_2 += nums[index2]
middle_right = max(middle_right, auxiliary_2)
middle = middle_left + middle_right
result = max(left, right, middle)
return result