> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/dynamic_programming_i/longest-continuous-increasing-subsequence.md).

# 674. Longest Continuous Increasing Subsequence

https\://leetcode.com/problems/longest-continuous-increasing-subsequence/description/

Given an unsorted array of integers, find the length of longest `continuous` increasing subsequence (subarray).

**Example 1:**<br>

```
Input: [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. 
Even though [1,3,5,7] is also an increasing subsequence, it's not a continuous one where 5 and 7 are separated by 4. 
```

**Example 2:**<br>

```
Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2], its length is 1. 
```

**Note:** Length of the array will not exceed 10,000.

## Thoughts

返回数组中最长的递增子数组长度。argmax + subarray => DP或dynamic window。以nums\[i]为底，action为是否与前面的拼接，dp存以nums\[i]为底最长长度。

## Code

```python
class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        if len(nums) == 0: return 0
        dp, res = 1, 1
        for i in range(1, len(nums)):
            dp = 1 if nums[i] <= nums[i - 1] else dp + 1
            res = max(res, dp)
        return res
```

```cpp
class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int max = 0;
        for (int i = 0, count = 0; i < nums.length; i++) {
            count++;
            if (i == nums.length - 1 || nums[i] >= nums[i + 1]) {
                max = Math.max(max, count);
                count = 0;
            } 
        }
        return max;
    }
}
```

## Analysis

时间复杂度O(N), 空间O(1).