README_EN.md

674. Longest Continuous Increasing Subsequence

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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].

 

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

Solutions

Python3

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        n = len(nums)
        if n < 2:
            return n
        res = f = 1
        for i in range(1, n):
            f = 1 + (f if nums[i - 1] < nums[i] else 0)
            res = max(res, f)
        return res

Java

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int n;
        if ((n = nums.length) < 2) return n;
        int res = 1, f = 1;
        for (int i = 1; i < n; ++i) {
            f = 1 + (nums[i - 1] < nums[i] ? f : 0);
            res = Math.max(res, f);
        }
        return res;
    }
}

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