Leetcode 5 Longest palindromic Substring (C,c++,python,java)

Problem:Given A stringS, find the longest palindromic substring inS. Assume that maximum length ofSis, and there exists one unique longest palindromic substring.
Solution:find palindrome strings in the middle of each character, then record the largest palindrome string, time complexity O (n^2)Main topic:returns the longest string of strings, given a string. Problem Solving Ideas:the easiest way to think of brute force, determine each substring, and then see if it is a palindrome string, time complexity O (n^3), easy to time out, using the method in solution, time to take charge of the grams reached O (n^2), is said to be able to passManacher's Algorithmalgorithm to increase the time complexity to O (n), but not quite understand the meaning of the original link: http://blog.csdn.net/han_xiaoyang/article/details/11969497#t20Java source code (spents 294ms, it's long enough):

public class Solution {public String longestpalindrome (String s) {int max=1,maxf=0,maxe=0;            for (int i=0;i<s.length (); i++) {int end = findodd (s,i);                if (Max < (end-i) *2+1) {max = (end-i) *2+1;                MAXF = I+i-end;            Maxe=end;            } end = Findeven (s,i);                if (Max < (end-i) * *) {max = (end-i) * *;                MAXF = I+i+1-end;            Maxe = end;    }} return S.substring (maxf,maxe+1);        } public int findodd (String s,int Center) {int i=center-1,j=center+1;            while (i>=0 && j<s.length ()) {if (S.charat (i)!=s.charat (j)) return j-1;        i--;j++;    } return j-1;        } public int Findeven (String s,int Center) {int i=center,j=center+1;            while (i>=0 && j<s.length ()) {if (S.charat (i)!=s.charat (j)) return j-1;        i--;j++; } return J1; }}

C Language Source code (spents 29ms):

int findodd (char* s,int Center) {    int i=center-1,j=center+1;    while (i>=0 && s[j]) {        if (S[i]!=s[j]) return j-1;        i--;j++;    }    return j-1;} int Findeven (char* s,int Center) {    int i=center,j=center+1;    while (i>=0 && s[j]) {        if (S[i]!=s[j]) {            return j-1;        }        i--;j++;    }    return j-1;} char* Longestpalindrome (char* s) {    int i=0,end,max=1,maxf=0,maxe=0;    for (i=0;s[i];i++) {        end=findodd (s,i);        if (max< (end-i) *2+1) {            max= (end-i) *2+1;            Maxf=i+i-end; maxe=end;        }        End=findeven (s,i);        if (max< (end-i) * *) {            max= (end-i) * *;            Maxf=i+i+1-end; maxe=end;        }    }    s[maxe+1]=0;    return S+MAXF;}

C + + source code (spents 95ms):

Class Solution {Public:string Longestpalindrome (string s) {int max=1,maxf=0,maxe=0;            for (int i=0;i<s.size (); i++) {int end = findodd (s,i);                if (Max < (end-i) *2+1) {max = (end-i) *2+1;                Maxf=i+i-end;            Maxe=end;            } end = Findeven (s,i);                if (Max < (end-i) * *) {max = (end-i) * *;                Maxf=i+i+1-end;            Maxe=end;    }} return S.substr (Maxf,max);        } int findodd (string S,int Center) {int i=center-1,j=center+1;            while (i>=0 && j<s.size ()) {if (S[i]!=s[j]) return j-1;        i--;j++;    } return j-1;        } int Findeven (string S,int Center) {int i=center,j=center+1;            while (i>=0 && j<s.size ()) {if (S[i]!=s[j]) return j-1;        i--;j++;    } return j-1; }};

python source code (spents 1434ms, this really too long, to improve the algorithm):

Class solution:    # @param {string} s    # @return {string}    def longestpalindrome (self, s):        max=1; maxf=0; Maxe=0 for        i in range (0,len (s)):            end = Self.findodd (s,i)            if Max < (end-i) *2+1:                max = (end-i) *2+1                MAXF = i+i-end                maxe = end            end = Self.findeven (s,i)            if Max < (end-i) * 2:                max = (end-i) * *                MAXF = i +i+1-end                maxe = end        return s[maxf:maxe+1]    def findodd (self,s,center):        i=center-1;j=center+1 While        I>=0 and J<len (s):            if S[i]!=s[j]:return j-1            i=i-1;j=j+1        return j-1    def Findeven (self,s,center):        i=center;j=center+1 while        i>=0 and J<len (s):            if S[i]!=s[j]:return j-1            i= I-1;j=j+1        return j-1

Leetcode 5 Longest palindromic Substring (C,c++,python,java)