A palindrome is a nonempty string over some alphabet that reads the same forward
and backward. Examples of palindromes are all strings of length 1, civic,
racecar, and aibohphobia (fear of palindromes).
Give an efficient algorithm to find the longest palindrome that is a subsequence
of a given input string. For example, given the input character, your algorithm
should return carac. What is the running time of your algorithm?

struct Palindrome {
std::string strPalindrome;
std::string strFront;
std::string strEnd;
}; using VectorOfVectorOfIntermediate = std::vector<std::vector<Palindrome>>; std::string findLongestPalindrome(const std::string &strInput) {
VectorOfVectorOfIntermediate vecVecIntermediate = VectorOfVectorOfIntermediate(strInput.size(), std::vector<Palindrome>(strInput.size() + ));
for ( int i = ; i < strInput.size(); ++ i ) {
vecVecIntermediate[i][].strPalindrome = strInput[i];
vecVecIntermediate[i][].strFront = "";
vecVecIntermediate[i][].strEnd = "";
} for ( int i = ; i < strInput.size() - ; ++ i ) {
if ( strInput[i] == strInput[i + ]) {
vecVecIntermediate[i][].strPalindrome = strInput.substr(i, );
vecVecIntermediate[i][].strFront = "";
vecVecIntermediate[i][].strEnd = "";
}else {
vecVecIntermediate[i][].strPalindrome = strInput[i];
vecVecIntermediate[i][].strEnd = strInput[i+];
}
} for ( int L = ; L <= strInput.size(); L += ) {
for ( int n = ; n <= strInput.size() - L; ++ n ) {
size_t findPos; Palindrome p2 = vecVecIntermediate[n][L-];
p2.strEnd.push_back ( strInput[n + L - ] );
findPos = p2.strFront.find_first_of ( p2.strEnd );
if ( findPos != std::string::npos) {
auto charFind = p2.strFront[findPos];
p2.strPalindrome.insert ( p2.strPalindrome.begin(), charFind );
p2.strPalindrome.push_back ( charFind );
p2.strFront = p2.strFront.substr ( , findPos );
findPos = p2.strEnd.find ( charFind );
p2.strEnd = p2.strEnd.substr ( findPos + );
} Palindrome p3 = vecVecIntermediate[n+][L-];
p3.strFront.insert ( p3.strFront.begin(), strInput[n] );
findPos = p3.strFront.find_first_of ( p3.strEnd );
if ( findPos != std::string::npos) {
auto charFind = p3.strFront[findPos];
p3.strPalindrome.insert ( p3.strPalindrome.begin(), charFind );
p3.strPalindrome.push_back ( charFind );
p3.strFront = p3.strFront.substr ( , findPos );
findPos = p3.strEnd.find ( charFind );
p3.strEnd = p3.strEnd.substr ( findPos + );
} std::vector<Palindrome> vecP{p2, p3};
int nMaxIndex = ;
for ( int index = ; index < vecP.size(); ++ index ) {
if ( vecP[index].strPalindrome.length() > vecP[nMaxIndex].strPalindrome.length() ) {
nMaxIndex = index;
}
}
vecVecIntermediate[n][L] = vecP[nMaxIndex];
}
}
return vecVecIntermediate[][strInput.size()].strPalindrome;
} void Test_findLongestPalindrome()
{
{
std::string strTestCase("a");
auto strPalindrome = findLongestPalindrome ( strTestCase );
std::cout << "Test case " << strTestCase << ", result " << strPalindrome << std::endl;
} {
std::string strTestCase("aa");
auto strPalindrome = findLongestPalindrome ( strTestCase );
std::cout << "Test case " << strTestCase << ", result " << strPalindrome << std::endl;
} {
std::string strTestCase("ab");
auto strPalindrome = findLongestPalindrome ( strTestCase );
std::cout << "Test case " << strTestCase << ", result " << strPalindrome << std::endl;
} {
std::string strTestCase("abbac");
auto strPalindrome = findLongestPalindrome ( strTestCase );
std::cout << "Test case " << strTestCase << ", result " << strPalindrome << std::endl;
} {
std::string strTestCase("abcdefghijkjhgfed");
auto strPalindrome = findLongestPalindrome ( strTestCase );
std::cout << "Test case " << strTestCase << ", result " << strPalindrome << std::endl;
} {
std::string strTestCase("character");
auto strPalindrome = findLongestPalindrome ( strTestCase );
std::cout << "Test case " << strTestCase << ", result " << strPalindrome << std::endl;
} {
std::string strTestCase("GEEKS FOR GEEKS");
auto strPalindrome = findLongestPalindrome ( strTestCase );
std::cout << "Test case " << strTestCase << ", result " << strPalindrome << std::endl;
}
}

Longest palindrome subsequence的更多相关文章

  1. Uva 11151 - Longest Palindrome

    A palindrome is a string that reads the same from the left as it does from the right. For example, I ...

  2. [LeetCode] 409. Longest Palindrome 最长回文

    Given a string which consists of lowercase or uppercase letters, find the length of the longest pali ...

  3. [LeetCode] 516. Longest Palindromic Subsequence 最长回文子序列

    Given a string s, find the longest palindromic subsequence's length in s. You may assume that the ma ...

  4. [LeetCode] Longest Palindrome 最长回文串

    Given a string which consists of lowercase or uppercase letters, find the length of the longest pali ...

  5. [LeetCode] Longest Increasing Subsequence 最长递增子序列

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

  6. 动态规划求最长公共子序列(Longest Common Subsequence, LCS)

    1. 问题描述 子串应该比较好理解,至于什么是子序列,这里给出一个例子:有两个母串 cnblogs belong 比如序列bo, bg, lg在母串cnblogs与belong中都出现过并且出现顺序与 ...

  7. [tem]Longest Increasing Subsequence(LIS)

    Longest Increasing Subsequence(LIS) 一个美丽的名字 非常经典的线性结构dp [朴素]:O(n^2) d(i)=max{0,d(j) :j<i&& ...

  8. [LintCode] Longest Increasing Subsequence 最长递增子序列

    Given a sequence of integers, find the longest increasing subsequence (LIS). You code should return ...

  9. LintCode Longest Common Subsequence

    原题链接在这里:http://www.lintcode.com/en/problem/longest-common-subsequence/ 题目: Given two strings, find t ...

随机推荐

  1. DES加密与解密控制台c++代码

    #include"stdafx.h" #include<stdio.h> #include<string.h> void main() { //声明变量 c ...

  2. Linux-Linux基础入门

    第一节 Linux系统简介 初步了解了什么是Linux系统,有何优势.与Windows系统有何不同,并了解Linux学习方法. 第二节 基本概念及概念 1.完成实验楼入门基础课程,共两个实验:(1)& ...

  3. Javascript 控制 让输入框不能输入 数字

    监听keypress事件.判断如果是数字的话阻止浏览器冒泡 <input type="text" id="test"> <script typ ...

  4. Canvas vs. SVG[转]

    Canvas 和 SVG 都允许您在浏览器中创建图形,但是它们在根本上是不同的. SVG SVG 是一种使用 XML 描述 2D 图形的语言. SVG 基于 XML,这意味着 SVG DOM 中的每个 ...

  5. Promises讲解

    原生 Promises 是在 ES2015 对 JavaScript 做出最大的改变.它的出现消除了采用 callback 机制的很多潜在问题,并允许我们采用近乎同步的逻辑去写异步代码. 可以说 pr ...

  6. 谷歌浏览器插件开发入门-官方版Helloworld详解

    目录: 需求 原理 实现步骤: 一个空的插件 一个可以设置一种背景色的插件(可以设置百度首页的背景色为绿色) 一个可以设置多种背景色的插件 需求: 插件可以改变特定网址的背景颜色. 原理: 将各种ht ...

  7. Java NIO学习-预备知识

    java NIO加入了Channels.Buffers.Selector.通过他们可以为java的io添加非阻塞IO. 一.对于经典java IO库 1.除了Buffered开头的类,其他均没有加缓冲 ...

  8. Delphi 按Esc快捷键退出程序的简单方法

     第一种方法: 在窗体上放一个按钮: 1>.设置按钮的Cancel属性为True: 2>.在按钮的点击事件中写: procedure TForm1.btn1Click(Sender: TO ...

  9. 使用pycharm专业版创建虚拟环境

    Location为工程地址 D:\My_python 第二个Location为 虚拟环境放在这个工程下 Base interpreter:基于那个解释器来创建虚拟环境 Create后进入到目录查看下

  10. mysql on duplicate key update 和 insert ignore into

    on duplicate key update <insert id="insert" parameterType="Plan"> insert i ...