> For the complete documentation index, see [llms.txt](https://nataliekung.gitbook.io/ladder_code/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://nataliekung.gitbook.io/ladder_code/l4dong-tai-gui-hua/longest-turbulent-subarray.md).

# Longest Turbulent Subarray

A subarray `A[i], A[i+1], ..., A[j]` of `A` is said to be *turbulent* if and only if:

* For `i <= k < j`, `A[k] > A[k+1]` when `k` is odd, and `A[k] < A[k+1]` when `k` is even;
* **OR**, for `i <= k < j`, `A[k] > A[k+1]` when `k` is even, and `A[k] < A[k+1]` when `k` is odd.

That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.

Return the **length** of a maximum size turbulent subarray of A.

**Example 1:**

```
Input: [9,4,2,10,7,8,8,1,9]
Output: 5
Explanation: (A[1] > A[2] < A[3] > A[4] < A[5])
```

**Example 2:**

```
Input: [4,8,12,16]
Output: 2
```

**Example 3:**

```
Input: [100]
Output: 1
```

**Note:**

1. `1 <= A.length <= 40000`
2. `0 <= A[i] <= 10^9`

分析：

若弯曲则up由down来，down由up来，这样其实一直保存了最大曲线的长度且递增，同时也保存了当前现状。

loop里每次res也更新得到最大。

```
class Solution:
    def maxTurbulenceSize(self, A: List[int]) -> int:
        n = len(A)
        up,down,res=1,1,1
        for i in range(1,n):
            if A[i] > A[i-1]:
                up = down+1
                down =1
            elif A[i] < A[i-1]:
                down = up+1
                up = 1
            else:
                up=down=1
            res = max(res,up,down)
        return res
```