✅ [Rust] 0 ms, linear scan (with detailed comments)
This solution employs a simple loop to count the number of characters in the longest continuous increasing subsequence. It demonstrated 0 ms runtime (100.00%) a
This [solution](https://leetcode.com/submissions/detail/798332889/) employs a simple loop to count the number of characters in the longest continuous increasing subsequence. It demonstrated **0 ms runtime (100.00%)** and used **2.0 MB memory (100.00%)**. Detailed comments are provided.\n\n**IF YOU LIKE THIS SOLUTION, PLEASE UPVOTE.**\n```\nimpl Solution \n{\n    pub fn find_length_of_lcis(nums: Vec<i32>) -> i32 \n    {\n        let mut max_len : i32 = 0;\n        let mut cur_len : i32 = 0;\n        let mut prev    : i32 = -1_000_000_001;\n        \n        // [1] we keep track of current length \'cur_len\' of subsequence\n\t\t//     and update \'max_len\' every time the continuous sequence is broken\n        for n in nums\n        {\n            if n > prev { cur_len += 1; }\n            else        { max_len = max_len.max(cur_len); cur_len = 1; }\n            prev = n;\n        }\n        \n        // [2] update \'max_len\' for the last continuous sequence\n        max_len.max(cur_len)\n    }\n}\n```
['Rust']