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:

  • 0 <= nums.length <= 104
  • -109 <= nums[i] <= 109

Rust Solution

struct Solution;

impl Solution {
    fn find_length_of_lcis(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        if n == 0 {
            return 0;
        }
        let mut i: usize = 0;
        let mut j: usize = i;
        let mut res: usize = 1;
        while i < n {
            while j + 1 < n && nums[j + 1] > nums[j] {
                j += 1;
            }
            res = usize::max(j - i + 1, res);
            i = j + 1;
            j = i;
        }
        res as i32
    }
}

#[test]
fn test() {
    let nums = vec![1, 3, 5, 4, 7];
    assert_eq!(Solution::find_length_of_lcis(nums), 3);
    let nums = vec![2, 2, 2, 2, 2];
    assert_eq!(Solution::find_length_of_lcis(nums), 1);
}

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