[LeetCode] 674. Longest Continuous Increasing Subsequence_Easy Dynamic Programming

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.

这个是一个经典的DP问题,  A[i] = A[i-1] + 1 if a[i] > a[i-1] else 1     (i > 0)   , init: A[0] = 1

是[LeetCode] 329. Longest Increasing Path in a Matrix_Hard tag: Dynamic Programming, DFS, Memoization的一个前身.

code

T; O(n)    S; O(1)

class Solution:

    def longestContinuesSubarry(self, nums):

        if not nums: return 0

        n = len(nums)

        dp, ans = [1]*2, 1

        for i in range(1, n):

            if nums[i] > nums[i-1]:

                dp[i%2] = dp[(i - 1) % 2] + 1

            else:

                dp[i%2] = 1

            ans = max(ans, dp[i%2])

        return ans