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.

题目地址: Longest Continuous Increasing Subsequence

难度: Easy

题意: 找出呈递增的趋势的子数组,返回最大长度

思路:

遍历数组,并计数,比较简单

class Solution(object):
def findLengthOfLCIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
n = len(nums)
if n <= 1:
return len(nums) res = 0
length = 1
for i in range(n-1):
if nums[i] < nums[i+1]:
length += 1
else:
res = max(length, res)
length = 1
res = max(length, res)
return res

时间复杂度: O(n)

空间复杂度: O(1)

674. Longest Continuous Increasing Subsequence@python的更多相关文章

  1. leetcode300. Longest Increasing Subsequence 最长递增子序列 、674. Longest Continuous Increasing Subsequence

    Longest Increasing Subsequence 最长递增子序列 子序列不是数组中连续的数. dp表达的意思是以i结尾的最长子序列,而不是前i个数字的最长子序列. 初始化是dp所有的都为1 ...

  2. 【Leetcode_easy】674. Longest Continuous Increasing Subsequence

    problem 674. Longest Continuous Increasing Subsequence solution class Solution { public: int findLen ...

  3. [LeetCode&Python] Problem 674. Longest Continuous Increasing Subsequence

    Given an unsorted array of integers, find the length of longest continuousincreasing subsequence (su ...

  4. 【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/ 目录 题目描述 题目大意 解题方法 动态规划 空间压缩DP 日期 题目地址:https: ...

  5. [LeetCode] 674. Longest Continuous Increasing Subsequence 最长连续递增序列

    Given an unsorted array of integers, find the length of longest continuous increasing subsequence. E ...

  6. LeetCode 674. Longest Continuous Increasing Subsequence (最长连续递增序列)

    Given an unsorted array of integers, find the length of longest continuous increasing subsequence. E ...

  7. [Leetcode]674. Longest Continuous Increasing Subsequence

    Given an unsorted array of integers, find the length of longest continuous increasing subsequence. E ...

  8. 674. Longest Continuous Increasing Subsequence最长连续递增子数组

    [抄题]: Given an unsorted array of integers, find the length of longest continuous increasing subseque ...

  9. LeetCode 674. Longest Continuous Increasing Subsequence最长连续递增序列 (C++/Java)

    题目: Given an unsorted array of integers, find the length of longest continuous increasing subsequenc ...

随机推荐

  1. Codevs 1293 送给圣诞夜的极光

    1293 送给圣诞夜的极光  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 黄金 Gold 题解  查看运行结果     题目描述 Description 圣诞老人回到了北极圣 ...

  2. CF939D Love Rescue

    题意 给定两个长度为n的由小写字母组成的字符串 每次可以花费1的代价,指定两个字母,把其中一个全部变为另一个 求使两个字符串相同的最小花费 n <= 100000 因为谁变成谁没有关系反正相等就 ...

  3. 关于Dictionary的优化用法

    今天突然想到了解一下Dictionary,于是在博客园上看到了一篇关于用TryGetValue的文章,原来用TryGetValue要比用ContainsKey更快,快一倍.

  4. 买票案例 1.synchronize关键字 2.lock锁

  5. mongodb vs redis(Tokyo Tyrant转)

    * MongoDB vs Redis vs Tokyo Tyrant(原文链接:http://www.cnblogs.com/riceball/archive/2010/03/05/MongoDB_V ...

  6. Python及bs4、lxml、numpy模块包的安装

    http://blog.csdn.net/tiantiancsdn/article/details/51046490(转载) Python及bs4.lxml.numpy模块包的安装 Python 的安 ...

  7. TabBar背景颜色设置

    // 第一种方式 // [[UITabBar appearance] setBarTintColor:[UIColor blackColor]]; // [UITabBar appearance].t ...

  8. volatile关键字简单摘要

    volatile就可以说是java虚拟机提供的最轻量级的同步机制 特性: 1.保证共享变量的可见性,即一个线程修改了某个变量的值,这新值对其他线程来说是立即可见的——要了解主存.高速缓存还有Java内 ...

  9. git stash暂存当前正在进行的工作

    git stash 可用来暂存当前正在进行的工作, 比如想pull 最新代码, 又不想加新commit, 或者另外一种情况,为了fix 一个紧急的bug,  先stash, 使返回到自己上一个comm ...

  10. (转)Quirks模式与standards模式区别

    建议:不推荐使用Quirks Mode. Quirks Mode中发生了什么?Quirks Mode是一种浏览器(像IE,Firefox,Opera)操作模式.从根本上说,怪异模式(也称之为兼容模式) ...