L2-008 longest substring symmetry (25 minutes)

L2-008 longest substring symmetry (25 minutes)

For a given string, this question requires you to output the length of the longest substring symmetrical. For example, given Is PAT&TAP symmetric?the longest substring of symmetry s PAT&TAP s, so you should output 11.

Input formats:

Input gives the length of not more than 1000 non-empty string in a row.

Output formats:

Symmetrical output longest substring of length in a row.

Sample input:

Is PAT&TAP symmetric?

Sample output:

11

 

Thinking: Manacher time complexity of O (n) solving palindromic longest substring. 

#include <iostream>  
#include <cstring>
#include <algorithm>   
using namespace std; 
string s; 
char s_new[4000];
int  p[5000]; 
int Init(){
    int len = s.length();
    s_new[0] = '$';
    s_new[1] = '#';
    int j = 2; 
    for (int i = 0; i < len; i++){
        s_new[j++] = s[i];
        s_new[j++] = '#';
    }
    s_new[j] = '\0';  // 别忘了哦
     return j;  // 返回 s_new 的长度
}
 
int Manacher(){
    int len = Init();  // 取得新字符串长度并完成向 s_new 的转换
    int max_len = -1;  // 最长回文长度
    int id;
    int mx = 0;
    for (int i = 1; i < len; i++){
        if (i < mx)
            p[i] = min(p[2 * id - i], mx - i);  // 需搞清楚上面那张图含义, mx 和 2*id-i 的含义
        else
            p[i] = 1;
 
        while (s_new[i - p[i]] == s_new[i + p[i]])  // 不需边界判断，因为左有'$',右有'\0'
            p[i]++;
 
        // 我们每走一步 i，都要和 mx 比较，我们希望 mx 尽可能的远，这样才能更有机会执行 if (i < mx)这句代码，从而提高效率
        if (mx < i + p[i]) {
            id = i;
            mx = i + p[i];
        }
        max_len = max(max_len, p[i] - 1);
    }
    return max_len;
}
 
int main(){ 
        
       getline(cin, s);
       printf("%d\n", Manacher()); 
    return 0;
}

 

 

Dynamic Programming:

      Analysis: There are two possibilities, ⼀ kind of back to the file is ⻓ string is odd, ⼀ species is an even number. the subscript i is the current character string.

                 When the back is an odd number when the file string, j denotes the i ij ⻓ string files back configuration of j +; ⻓ string when the back of the file is an even number, j represents

                 i + 1 j characters left ⼀ back to the file of the string ⻓ the right until i j characters ~

                 Use maxvalue, the maximum values ​​and save the resulting output traversal ~

 

#include <iostream>
using namespace std;
int main() {
 string s;
 getline(cin, s);
 int maxvalue = 0, temp;
 int len = s.length();
 for( int i = 0; i < len; i++) {
      temp = 1;
      for( int j = 1; j < len; j++) {
           if( i - j < 0 || i + j >= len || s[i - j] != s[i + j])
               break;
           temp += 2;
       }
      maxvalue = temp > maxvalue ? temp : maxvalue;
      temp = 0;
      for( int j = 1; j < len; j++) {
           if( i - j + 1 < 0 || i + j >= len || s[i - j + 1] != s[i + j])
              break;
           temp += 2;
      }
     maxvalue = temp > maxvalue ? temp : maxvalue;
 }
 cout << maxvalue;
 return 0;
}