Leetcode最大和最小子序和

最大子序和，动态规划时间复杂度为O(n)，则假设每一步为step，
$$sf(x_1)=x_1$$

$$sf(x_2)=\{\begin{array}{l}sf(x_1)+x_2,sf(x_1)+x_2>x_2\\ x_2,sf(x_1)+x_2<=x_2\end{array}$$

$$sf(x_3)=\{\begin{array}{l}sf(x_2)+x_3,sf(x_2)+x_3>x_3\\ x_3,sf(x_2)+x_3<=x_3\end{array}$$

$$f(x_1,x_2)=max(sf(x_1),sf(x_2))$$

$$f(x_1,x_2,x_3)=max(sf(x_1),sf(x_2),sf(x_3))$$

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        if len(nums) == 0:
            return 0

        global_max = nums[0]
        step_max = nums[0]

        if len(nums) == 1:
            return global_max

        for i in range(1, len(nums)):
            if step_max + nums[i] < nums[i]:
                step_max = nums[i]
            else:
                step_max += nums[i]

            if step_max > global_max:
                global_max = step_max

        return global_max

class Solution:
    def minSubArray(self, nums: List[int]) -> int:
        if len(nums) == 0:
            return 0

        global_min = nums[0]
        step_min = nums[0]

        if len(nums) == 1:
            return global_min

        for i in range(1, len(nums)):
            if step_min + nums[i] < nums[i]:
                step_min += nums[i]
            else:
                step_max = nums[i]

            if step_min < global_min:
                global_min = step_min

        return global_min

https://www.zhihu.com/question/23995189/answer/35324479
https://www.zhihu.com/question/52165201