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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0601_0700.s0674_longest_continuous_increasing_subsequence;

// #Easy #Array #2022_03_22_Time_2_ms_(40.35%)_Space_45.9_MB_(35.43%)

/**
 * 674 - Longest Continuous Increasing Subsequence\.
 *
 * Easy
 *
 * 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
**/
public class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int ans = 1;
        int count = 1;
        for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i] < nums[i + 1]) {
                count++;
            } else {
                ans = max(count, ans);
                count = 1;
            }
        }
        return max(ans, count);
    }

    private int max(int n1, int n2) {
        return Math.max(n1, n2);
    }
}