# 674. Longest Continuous Increasing Subsequence

Given an unsorted array of integers `nums`, return *the length of the longest **continuous increasing subsequence** (i.e. subarray)*. The subsequence must be **strictly** increasing.

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:**

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

**Example 2:**

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

**Constraints:**

* `1 <= nums.length <= 104`
* `-109 <= nums[i] <= 109`

***My Solutions:***

```java
public int findLengthOfLCIS(int[] nums) {
    if (nums == null || nums.length == 0) return 0;
    int max = 1, index = 0; // index 表示连续递增的第一个数字所在的位置
    
    for (int i = 1; i < nums.length; i++) {
        if (nums[i] > nums[i - 1]) {
            max = Math.max(max, i - index + 1);
        } else {
            index = i;
        }
    }
    return max;
}
```
