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alogo001.tar
- C++ algorithm: 1) 10001th prime number 2) Number that divisible fro, 1 to 20 3) Biggest palindromic number for two 3-digit number s product
zhanduiliesuanfa
- 一、数制转换问题；二、回文判断；三、括号匹配的检验。-A number system conversion two palindromic judge three, bracket matching test.
Manacher
- Manacher 算法， JAVA描述 Computes the longest palindromic substring in linear time using Manacher s algorithm.-Manacher Algorithm implemented by Java Computes the longest palindromic substring in linear time using Manacher s algorithm.
key_Queue
- 3． 参考void Reverse(LinkQueue &Q)函数，定义新的函数int Palindrome_Test(char str[])，判断字符串str是否回文序列,若是则返回1,否则返回0。写出程序或用文字描述算法的执行步骤。-3. Reference void Reverse (LinkQueue & Q) function, define a new function int Palindrome_Test (char str []), to determine wheth
palindromic-numbers
- Program to find palindromic-numbers
getAllPalindromeSubStr
- 给定字符串，求它的回文子序列个数。回文子序列反转字符顺序后仍然与原序列相同。例如字符串aba中，回文子序列为 a , a , aa , b , aba ，共5个。内容相同位置不同的子序列算不同的子序列。-Given a string, find its number of sub-palindromic sequence. Palindromic sequences reversed the order of characters remains the same as the original
Longest-Palindromic-Substring
- 这是一个求最长回纹序列的c++算法，简单易懂，感兴趣的可以-This is a request fret longest sequence c++ algorithm, easy to understand, are interested can look at
Prime-statistic
- 经过优化省时的素数统计，素数回文，统计因子和的源码，值得初学者学习。-After primes optimized saving statistics, palindromic primes, statistical factors and source code, it is worth beginners to learn.
5-Longest-Palindromic-Substring
- “回文串”是一个正读和反读都一样的字符串，比如“level”或者“noon”等等就是回文串。回文子串，顾名思义，即字符串中满足回文性质的子串。 Manacher算法- Palindrome string is a positive reading and verlan are the same string, such as level or noon and so is a palindrome string. Palindrome substring, by definitio
Manacher_Longest_Subsequence
- 求一个字符串的最长回文子串的长度。比如字符串 abaaba 的最长回文子串的长度就是6.-Seeking a string longest palindromic substring length. For example, the string " abaaba" longest palindromic substring length is 6.