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:

Input: nums = [1,3,5,4,7]
Output: 3
Explanation: 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: 1
Explanation: The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly
increasing.

Constraints:

• 1 <= nums.length <= 104

• -109 <= nums[i] <= 109

My Solutions:

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;
}