674. Longest Continuous Increasing Subsequence

1. Description

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].

2. Example

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.

3. Constraints

• 0 <= nums.length <= 10$^{4}$
• -10$^{9}$ <= nums[i] <= 10$^{9}$

4. Solutions

Array

n = i_nums.size()
Time complexity: O(n)
Space complexity: O(1)

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class Solution
{
public:
    // int findLengthOfLCIS(vector<int>& nums)
    int findLengthOfLCIS(vector<int> &i_nums)
    {
        if(i_nums.empty()) return 0;

        int result = 1;
        for(int i = 1, length = 1; i < i_nums.size(); i++)
        {
            length = (i_nums[i] > i_nums[i - 1] ? length + 1 : 1);

            result = max(result, length);
        }

        return result;
    }
};