Given an unsorted array of integers `nums`

, return *the length of the longest continuous increasing subsequence (i.e. subarray)*. The subsequence must be

A **continuous increasing subsequence** is defined by two indices `l`

and `r`

(`l < r`

) such that it is `[nums[l], nums[l + 1], ..., nums[r - 1], nums[r]]`

and for each `l <= i < r`

, `nums[i] < nums[i + 1]`

.

**Example 1:**

Input:nums = [1,3,5,4,7]Output:3Explanation:The longest continuous increasing subsequence is [1,3,5] with length 3. Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element 4.

**Example 2:**

Input:nums = [2,2,2,2,2]Output:1Explanation:The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly increasing.

**Constraints:**

`0 <= nums.length <= 10`

^{4}`-10`

^{9}<= nums[i] <= 10^{9}

```
struct Solution;
impl Solution {
fn find_length_of_lcis(nums: Vec<i32>) -> i32 {
let n = nums.len();
if n == 0 {
return 0;
}
let mut i: usize = 0;
let mut j: usize = i;
let mut res: usize = 1;
while i < n {
while j + 1 < n && nums[j + 1] > nums[j] {
j += 1;
}
res = usize::max(j - i + 1, res);
i = j + 1;
j = i;
}
res as i32
}
}
#[test]
fn test() {
let nums = vec![1, 3, 5, 4, 7];
assert_eq!(Solution::find_length_of_lcis(nums), 3);
let nums = vec![2, 2, 2, 2, 2];
assert_eq!(Solution::find_length_of_lcis(nums), 1);
}
```