5. Longest Palindromic Substring

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

本题求最长的回文子串，最简单的就是暴力求解，用中心扩展法，但是回文串有两种形式，奇abc和偶abba，两种情况都要考虑到。

#include<stdio.h>#include<string.h>int findodd(char *s,int i)//aba{    int j;    for(j=0;i-j>=0&&s[i+j];j++)    {        if(s[i-j]!=s[i+j])            break;    }    return j;}int findeven(char *s,int i)//abba{    int j;    for(j=0;i-j>=1&&s[i+j];j++)//开始这里我用的时i+j<strlen(s)，结果时间有150多ms，用s[i+j]减到24ms，所以要慎用字符串函数啊啊啊    {        if(s[i+j]!=s[i-j-1])            break;    }    return j;}char* longestPalindrome(char* s) {    int i,j,max,maxl,maxr;    max=0;    maxl=0;//回文子串左边开始位置    maxr=0;//回文子串右边结束位置    for(i=0;s[i];i++)    {        j=findodd(s,i);        if((2*j-1)>max)        {            max=2*j-1;            maxl=i-j+1;            maxr=i+j-1;        }        j=findeven(s,i);        if(2*j>max)        {            max=2*j;            maxl=i-j;            maxr=i+j-1;        }    }    s[maxr+1]='\0';    return s+maxl;}int main(){    char s[100];    scanf("%s",s);    printf("%s\n",longestPalindrome(s));     return 0;}

C++代码

class Solution {  public:      string str_helper(string s, int left, int right)      {          int n = s.length();          while(left>=0 && right<=n-1 && s[left]==s[right]){              left--;              right++;          }          return s.substr(left+1, right-left-1);      }      string longestPalindrome(string s) {              if(s.length() <= 1)                  return s;          string longest = s.substr(0,1);          for (int i=0; i<s.length(); i++){              string tmp = str_helper(s, i, i);              if (tmp.length() > longest.length())  {                  longest = tmp;              }              tmp = str_helper(s, i, i+1);              if (tmp.length() > longest.length()) {                  longest = tmp;              }          }          return longest;      }  };