[LeetCode] 674. Longest Continuous Increasing Subsequence_Easy Dynamic Programming
2018-08-02 09:18 Johnson_强生仔仔 阅读(233) 评论(0) 编辑 收藏 举报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