题目:

Given a sequence of integers, find the longest increasing subsequence (LIS).

You code should return the length of the LIS.

Clarification

What's the definition of longest increasing subsequence?

  • The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.

  • https://en.wikipedia.org/wiki/Longest_increasing_subsequence

Example

For [5, 4, 1, 2, 3], the LIS is [1, 2, 3], return 3
For [4, 2, 4, 5, 3, 7], the LIS is [2, 4, 5, 7], return 4

题解:

  For dp[i], dp[i] is max(dp[j]+1, dp[i]), for all j < i and nums[j] < nums[i].

Solution 1 ()

class Solution {
public:
int longestIncreasingSubsequence(vector<int> nums) {
if (nums.empty()) {
return ;
}
vector<int> dp(nums.size(), );
int res = ;
for (int i = ; i < nums.size(); ++i) {
for (int j = ; j < i; ++j) {
if (nums[j] < nums[i]) {
dp[i] = max(dp[i], dp[j] + );
}
}
res = max(dp[i], res);
}
return res;
}
};

Solution 2 ()

class Solution {
public:
/**
* @param nums: The integer array
* @return: The length of LIS (longest increasing subsequence)
*/
int longestIncreasingSubsequence(vector<int> nums) {
vector<int> res;
for(int i=; i<nums.size(); i++) {
auto it = std::lower_bound(res.begin(), res.end(), nums[i]);
if(it==res.end()) res.push_back(nums[i]);
else *it = nums[i];
}
return res.size();
}
};

Solution 3 ()

class Solution {
public:
int longestIncreasingSubsequence(vector<int> nums) {
if (nums.empty()) {
return ;
}
vector<int> tmp;
tmp.push_back(nums[]);
for (auto num : nums) {
if (num < tmp[]) {
tmp[] = num;
} else if (num > tmp.back()) {
tmp.push_back(num);
} else {
int begin = , end = tmp.size();
while (begin < end) {
int mid = begin + (end - begin) / ;
if (tmp[mid] < num) {
begin = mid + ;
} else {
end = mid;
}
}
tmp[end] = num;
}
}
return tmp.size();
}
};

【Lintcode】076.Longest Increasing Subsequence的更多相关文章

  1. 【leetcode】300.Longest Increasing Subsequence

    Given an unsorted array of integers, find the length of longest increasing subsequence. For example, ...

  2. 【LeetCode】300. Longest Increasing Subsequence 解题报告(Python & C++)

    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 日期 题目地址:https://leetcode.c ...

  3. 【Lintcode】077.Longest Common Subsequence

    题目: Given two strings, find the longest common subsequence (LCS). Your code should return the length ...

  4. 【题解】Greatest Common Increasing Subsequence

    [题解]Greatest Common Increasing Subsequence vj 唉,把自己当做DP入门选手来总结这道题吧,我DP实在太差了 首先是设置状态的技巧,设置状态主要就是要补充不漏 ...

  5. 【LeetCode】522. Longest Uncommon Subsequence II 解题报告(Python)

    [LeetCode]522. Longest Uncommon Subsequence II 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemin ...

  6. 【刷题-LeetCode】300. Longest Increasing Subsequence

    Longest Increasing Subsequence Given an unsorted array of integers, find the length of longest incre ...

  7. 【Leetcode_easy】594. Longest Harmonious Subsequence

    problem 594. Longest Harmonious Subsequence 最长和谐子序列 题意: 可以对数组进行排序,那么实际上只要找出来相差为1的两个数的总共出现个数就是一个和谐子序列 ...

  8. 【leetcode】521. Longest Uncommon Subsequence I

    problem 521. Longest Uncommon Subsequence I 最长非共同子序列之一 题意: 两个字符串的情况很少,如果两个字符串相等,那么一定没有非共同子序列,反之,如果两个 ...

  9. 【leetcode】1218. Longest Arithmetic Subsequence of Given Difference

    题目如下: Given an integer array arr and an integer difference, return the length of the longest subsequ ...

随机推荐

  1. JavaWeb学习总结第四篇--Servlet开发

    Servlet开发 用户在浏览器中输入一个网址并回车,浏览器会向服务器发送一个HTTP请求.服务器端程序接受这个请求,并对请求进行处理,然后发送一个回应.浏览器收到回应,再把回应的内容显示出来.这种请 ...

  2. eclipse集成tomcat改动字符集參数

    问题: 在eclipse 4.4(Luna)中集成tomcat时,直接改动原tomcat文件夹中的配置文件,不起作用. 有时.我们会修改字符集參数为utf-8,以解决中文乱码问题,修改之后依旧乱码-- ...

  3. Python 深入剖析SocketServer模块(一)(V2.7.11)

    一.简介(翻译)  通用socket server 类  该模块尽力从各种不同的方面定义server:  对于socket-based servers:  -- address family:     ...

  4. EasyPlayerPro(Windows)流媒体播放器开发之跨语言调用

    下面我们来讲解一下关于EasyPlayerPro接口的调用,主要分为C++和C#两种语言,C++也可以基于VC和QT进行开发,C++以VC MFC框架为例进行讲解,C#以Winform框架为例进行讲解 ...

  5. 一些重要的地址:md5在线解密破解

    md5在线解密破解:https://www.cmd5.com/

  6. 九度OJ 1035:找出直系亲属 (二叉树、递归)

    时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:2380 解决:934 题目描述:     如果A,B是C的父母亲,则A,B是C的parent,C是A,B的child,如果A,B是C的(外) ...

  7. Android开发之ListView添加多种布局效果演示

    在这个案例中展示的新闻列表,使用到ListView控件,然后在适配器中添加多种布局效果,这里通过重写BaseAdapter类中的 getViewType()和getItemViewType()来做判断 ...

  8. oracle 11g ocr 冗余配置

    版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/royjj/article/details/30506343  oracle 11g ocr 冗余 ...

  9. PYTHON调用C接口(基于Ctypes)实现stein算法最大公约数的计算

    相关环境配置 mingw,选择相应的32位.64位的版本,主要用于编译动态链接库dll文件,可用vs替代,这里我选择轻量级的mingw windows64位地址:https://sourceforge ...

  10. JS性能优化——DOM编程

    浏览器中的DOM  天生就慢 DOM是个与语言无关的API,它在浏览器中的接口却是用JavaScript实现的.客户端脚本编程大多数时候是在个底层文档打交道,DOM就成为现在JavaScript编码中 ...