674. Longest Continuous Increasing Subsequence

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

Submitted Code

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        count = 1       # subsequence 길이
        max_len = 1     # 가장 긴 subsequence

        for i in range(1, n):
            if nums[i] - nums[i-1] <= 0:
                max_len = max(count, max_len)
                count = 1
            else:
                count += 1
        
        return max(count, max_len)

Runtime: 3 ms | Beats 68.31%
Memory: 13.32 MB | Beats 60.24%

반복문 종료 후 리스트의 마지막 원소까지 신경써서 처리해줘야 한다.

Other Solutions

1st

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 0:
            return 0
        if len(nums) == 1:
            return 1

        left, right = 0, 1
        max_len = 1

        while right < len(nums):
            if nums[right] > nums[right - 1]:
                max_len = max(max_len, right - left + 1)
            else:
                left = right
            right += 1
        return max_len

time complexity: 𝑂(𝑛)
space complexity: 𝑂(1)

while문으로도 풀 수 있다.