[Leetcode] 005. Longest palindromic Substring (Medium) (C++/java/python)

Index: [Leetcode] leetcode key index (C++/JAVA/PYTHON/SQL)
Github:https://github.com/illuz/leetcode

005.longest_palindromic_substring (Medium) links：

Title: https://oj.leetcode.com/problems/Longest-Palindromic-Substring/
Code (GitHub): Https://github.com/illuz/leetcode

Test Instructions：

The longest palindrome substring in a string.

Analysis：

There are many solutions to palindrome:

1. Violent search O (n^3)
2. Dynamic planning O (n^2),dp[i][j] = dp[i + 1][j - 1] (if s[i] == s[j])
3. O (n) time can be reached with Manacher ' s algorithm.

The third algorithm is used in this case.
It is important to note that Python and Java strings are not the same as C + +, and do not \0 end with ' manacher ' algorithm '.

Code：

C++:

Class Solution {Public:string Longestpalindrome (string s) {int p[n<<1];string t = "$"; for (char ch:s) {T + = ' # '; t + = ch;} T + = ' # ';//T is a processed string, p is an array of record lengths  memset (p, 0, sizeof (p));  MX is the rightmost position of the determined palindrome, ID is the middle position, Mmax records the maximum value in the P array  int mx = 0, id = 0, Mmax = 0;  int len = T.length (), int right = 0;for (int i = 1; i < Len; i++) {  p[i] = mx > I min (p[2 * id-i], mx-i): 1;  while (T[i + p[i]] = = T[i-p[i]])  p[i]++;  if (i + p[i] > mx) {  mx = i + p[i];  id = i;  }  if (Mmax < p[i]) {Mmax = P[i]; right = i;}}  The longest is mmax-1  return s.substr (RIGHT/2-MMAX/2, mmax-1);}};

Java:

public class Solution {public String longestpalindrome (String s) {int[] p = new int[2048];        StringBuilder t = new StringBuilder ("$");            for (int i = 0; i < s.length (); ++i) {t.append (' # ');        T.append (S.charat (i));        } t.append ("#_");        MX is the rightmost position of the determined palindrome, ID is the middle position, Mmax records the maximum value in the p array int mx = 0, id = 0, Mmax = 0;        int right = 0; for (int i = 1; i < T.length ()-1; i++) {p[i] = mx > I?            Math.min (p[2 * id-i], mx-i): 1;            while (T.charat (i + p[i]) = = T.charat (I-p[i])) p[i]++;                if (i + p[i] > mx) {mx = i + p[i];            id = i;                } if (Mmax < p[i]) {Mmax = P[i];            right = i;    }}//Longest is mmax-1 return s.substring (RIGHT/2-MMAX/2, RIGHT/2-MMAX/2 + mmax-1); }}

Python:

Class solution:    # @return A string    def longestpalindrome (self, s):        t = ' $# ' + ' # '. Join (s) + ' #_ '        p = [0] * 4010        mx, id, mmax, right = 0, 0, 0, 0 for        I in range (1, len (t)-1):            if mx > I:                p[i] = min (p[2 * ID-  I], mx-i)            else:                p[i] = 1 while            t[i + p[i]] = = T[i-p[i]]:                p[i] + = 1            if i + p[i] > mx:                mx  = i + p[i]                id = i            if Mmax < P[i]:                Mmax = p[i] Right                = i        return S[RIGHT//2-MMAX//2:RIGHT//2 -MMAX//2 + mmax-1]

[Leetcode] 005. Longest palindromic Substring (Medium) (C++/java/python)