【LeetCode】674. Longest Continuous Increasing Subsequence 解题报告（Python）

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目描述

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.

题目大意

找出数组中最长连续递增子序列（子数组）

解题方法

动态规划

直接使用dp作为到某个位置的最长连续子序列。所以，如果当前的值比前一个值大，那么dp应该是前面的一个位置的数值+1，否则当前的值应该是1。另外需要注意的是当输入是空的时候，应该返回0.

class Solution(object):

    def findLengthOfLCIS(self, nums):

        """

        :type nums: List[int]

        :rtype: int

        """

        if not nums: return 0

        N = len(nums)

        dp = [1] * N

        for i in range(1, N):

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

                dp[i] = dp[i - 1] + 1

        return max(dp)

空间压缩DP

在上面的做法中看出，每步的结果之和上面一步有关，所以可以优化空间复杂度。

class Solution(object):

    def findLengthOfLCIS(self, nums):

        """

        :type nums: List[int]

        :rtype: int

        """

        longest = 0

        cur = 0

        for i in range(len(nums)):

            if i == 0 or nums[i] > nums[i - 1]:

                cur += 1

                longest = max(longest, cur)

            else:

                cur = 1

        return longest

二刷的时候版本。

class Solution(object):

    def findLengthOfLCIS(self, nums):

        """

        :type nums: List[int]

        :rtype: int

        """

        if not nums: return 0

        N = len(nums)

        dp = 1

        res = 1

        for i in range(1, N):

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

                dp += 1

                res = max(res, dp)

            else:

                dp = 1

        return res

日期

2018 年 1 月 29 日
2018 年 11 月 19 日 —— 周一又开始了

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