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:

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

Solutions

Python3

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        res, n = 1, len(nums)
        i = 0
        while i < n:
            j = i + 1
            while j < n and nums[j] > nums[j - 1]:
                j += 1
            res = max(res, j - i)
            i = j
        return res

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        n = len(nums)
        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 res = 1;
        for (int i = 1, f = 1; i < nums.length; ++i) {
            f = 1 + (nums[i - 1] < nums[i] ? f : 0);
            res = Math.max(res, f);
        }
        return res;
    }
}

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int res = 1;
        for (int i = 0, n = nums.length; i < n;) {
            int j = i + 1;
            while (j < n && nums[j] > nums[j - 1]) {
                ++j;
            }
            res = Math.max(res, j - i);
            i = j;
        }
        return res;
    }
}

C++

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        int res = 1;
        for (int i = 1, f = 1; i < nums.size(); ++i) {
            f = 1 + (nums[i - 1] < nums[i] ? f : 0);
            res = max(res, f);
        }
        return res;
    }
};

Go

func findLengthOfLCIS(nums []int) int {
	res, f := 1, 1
	for i := 1; i < len(nums); i++ {
		if nums[i-1] < nums[i] {
			f += 1
			res = max(res, f)
		} else {
			f = 1
		}
	}
	return res
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

TypeScript

function findLengthOfLCIS(nums: number[]): number {
    const n = nums.length;
    let res = 1;
    let i = 0;
    for (let j = 1; j < n; j++) {
        if (nums[j - 1] >= nums[j]) {
            res = Math.max(res, j - i);
            i = j;
        }
    }
    return Math.max(res, n - i);
}

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