pat 中求解最长回文串的长度

本以为暴力枚举会超时，但竟然过了。。。

#include <string>

#include <stdio.h>

#include <iostream>

int longestPalindrome(string s) {
int left = 0, right = s.length() - 1;
//为奇数时
int len1 = 0;
int len2 = 0;
int start1 = left;//标记起始点
int start2 = left;
for (int i = 0; i < s.length(); i++)
{
left = i;
right = i;
while (left >= 0 && right<s.length())
{
if (s[left] == s[right])
{
if (right - left + 1>len1)
{
len1 = right - left + 1;
start1 = left;
}
left--;
right++;
}
else
{
break;//重新找
}
}
}
//为偶数长度时
for (int i = 0; i < s.length() - 1; i++)
{
left = i;
right = i + 1;
while (left >= 0 && right<s.length())
{
if (s[left] == s[right])
{
if (right - left + 1>len2)
{
len2 = right - left + 1;
start2 = left;
}
left--;
right++;
}
else
{
break;
}
}
}
return len1> len2 ? len1 : len2;

}

int main()

{
//scanf("%s", s);
string s;
getline(cin, s);
int len = longestPalindrome(s);
cout << len << endl;
return 0;

}