[LeetCode] 674. Longest Continuous Increasing Subsequence_Easy Dynamic Programming

最新推荐文章于 2025-04-19 23:39:58 发布

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最新推荐文章于 2025-04-19 23:39:58 发布

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原文链接：http://www.cnblogs.com/Johnsonxiong/p/9405149.html

本文介绍了一种解决寻找最长连续递增子数组问题的方法，给出了一段O(n)时间复杂度和O(1)空间复杂度的Python代码实现。通过对输入数组的遍历，并利用动态规划的思想，有效地找到了最长连续递增子数组的长度。

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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