1040. Longest Symmetric String (25)

Given a string, you are supposed to output the length of the longest symmetric sub-string. For example, given “Is PAT&TAP symmetric?”, the longest symmetric sub-string is “s PAT&TAP s”, hence you must output 11.

Input Specification:

Each input file contains one test case which gives a non-empty string of length no more than 1000.

Output Specification:

For each test case, simply print the maximum length in a line.

Sample Input:

Is PAT&TAP symmetric?

Sample Output:

11

算法分析：由于数据比较小，采用穷举算法，还可以用动态规划。
1. 对于字符串中的每个字符的下标s[i]，首先考虑s[i]是否等于s[i - 1]，如果相等则继续比较s[i - 2]和s[i + 1];
2. 然后考虑s[i - 1]是否等于s[i + 1]，如果相等则继续比较s[i - 2]和s[i + 2]；
3. 在进行比较的同时还要更新回文串的长度。

这里举出一种动态规划算法，但是感觉计算复杂度都一样为O(n2)。传送门。

#include <stdio.h>#include <stdlib.h>#include <string.h>int main(){  int i, maxlen = 1;  char s[1001] = { '\0' };  gets(s);  int len = strlen(s);  for (i = 1; i < len; i++)  {    int front, tail, cl = 0;    if (s[i] == s[i - 1])    {      cl = 0;       //比较s[i-1]和s[i]      for (front = i - 1, tail = i; front >= 0 && tail < len; front--, tail++)        if (s[front] == s[tail])          cl += 2;        else          break;    }    else if (s[i - 1] == s[i + 1])    {      cl = 1; //注意这里初值为1      //比较s[i-1]和s[i+1]      for(front = i - 1, tail = i + 1; front >= 0 && tail < len; front--, tail++)        if (s[front] == s[tail])          cl += 2;        else          break;    }    if (cl > maxlen)      maxlen = cl;  }  printf("%d\n", maxlen);  system("pause");  return 0;}