674. Longest Continuous Increasing Subsequence

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本文探讨了如何找出未排序整数数组中最长连续递增子序列的长度，并提供了两种解决方案：一种使用动态规划，另一种通过遍历数组并维护最大长度来解决。这两种方法都有效地解决了问题。

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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.
1.动态规划：
dp[i] 表示当前位置的最大连续长度
公式：dp[i] = dp[i - 1] + 1 if (前一元素小于当前元素)

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        dp = [1]*len(nums)
        tmax=1
        for i in range(1,len(nums)):
            if nums[i]>nums[i-1]:
                dp[i]=dp[i-1]+1
            tmax = max(tmax,dp[i])
        return tmax

2.维护一个maxLen，每次遇到递增数字就tempLen++，遇到一个不是递增数字的话，就把tempLen 和maxLen 中大的保存到maxLen。

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        maxlen=1
        templen=1
        for i in range(1,len(nums)):
            if nums[i]>nums[i-1]:
                templen+=1
                maxlen=max(maxlen,templen)
            else:
                templen=1
        return maxlen