674. Longest Continuous Increasing Subsequence

Problem

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].

https://leetcode.com/problems/longest-continuous-increasing-subsequence/

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

Test Cases

1
2

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:

1
2
3
4
5
6
7
8
9
10
11
12

import pytest

from solution import Solution


@pytest.mark.parametrize('nums, expected', [
    ([1,3,5,4,7], 3),
    ([2,2,2,2,2], 1),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, nums, expected):
    assert sol.findLengthOfLCIS(nums) == expected

Thoughts

直接遍历一次数组，记录当前连续单调递增的子数组长度，并记录最大长度。

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

class Solution:
    def findLengthOfLCIS(self, nums: list[int]) -> int:
        max_len = 1
        prev = nums[0]
        curr_len = 1
        for num in nums:
            if num > prev:
                curr_len += 1
                if curr_len > max_len:
                    max_len = curr_len
            else:
                curr_len = 1

            prev = num

        return max_len

许可协议

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。

分享文章