674. Longest Continuous Increasing Subsequence

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

Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).

Example 1:

Input: [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. 
Even though [1,3,5,7] is also an increasing subsequence, it's not a continuous one where 5 and 7 are separated by 4. 

Example 2:

Input: [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2], its length is 1. 

Note: Length of the array will not exceed 10,000.

Thoughts

返回数组中最长的递增子数组长度。argmax + subarray => DP或dynamic window。以nums[i]为底，action为是否与前面的拼接，dp存以nums[i]为底最长长度。

Code

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

class Solution {
    public int findLengthOfLCIS(int[] nums) {
        int max = 0;
        for (int i = 0, count = 0; i < nums.length; i++) {
            count++;
            if (i == nums.length - 1 || nums[i] >= nums[i + 1]) {
                max = Math.max(max, count);
                count = 0;
            } 
        }
        return max;
    }
}

Analysis

时间复杂度O(N), 空间O(1).