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.

由于subarray is continous, 所以这题还是很简单的。

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