2017上海金马五校程序设计竞赛 E：Find Palindrome

Time Limit: 1 s

Description

Given a string S, which consists of lowercase characters, you need to find the longest palindromic sub-string.
A sub-string of a string S is another string S' that occurs "in" S. For example, "abst" is a sub-string of "abaabsta". A palindrome is a sequence of characters which reads the same backward as forward.

Input

There are several test cases. Each test case consists of one line with a single string S (1 ≤ |S | ≤ 50).

Output

For each test case, output the length of the longest palindromic sub-string.

Sample Input

sasadasa
bxabx
zhuyuan

Sample Output

7
1
3

题目意思：

给你一串字符，让你找出其中最长的回文子串长度。

解题思路：

首先先看字符串，对于第 i 个字符str[i]，如果回文子串有奇数个字符，那么以 i 为中心向左向右扩展，直到发现对称位置字符不相等，如果扫过X个字符，那么回文子串长度为2*X+1.

如果回文子串有偶数个字符，对于第 i 个字符str[i]在最接近对称轴的左边。对于这种情况我们要从第 i 个位置开始扫描，并与第 i+1 个位置字符进行比较，在向左向右扩展，直到对称位置字符不相同为止。如果扫过了X个字符，那么回文子串长度为2*X。

#include<iostream>  
#include<stdio.h>  
#include<algorithm>  
#include<string.h>  
#define INF 999999  
using namespace std;     
int main()  
{  
    int maxlen,i,j,len;   ///最长对称子串长度。  
    char str[1003];  
    while(gets(str))  
    {  
        maxlen = 0;  
        len = strlen(str);  
        for(i = 0; i < len; i++)  
        {  
            ///考虑是i是奇数串的中心,以i为中心，同时往左往右扩展  
            for(j = 0; i-j>=0&&i+j<=len; j++)  
            {  
                if(str[i-j]!=str[i+j])  
                    break;  
                if(2*j+1>maxlen)  
                    maxlen = 2*j+1;  
            }  
            ///i是偶数串的中心  
            for(j = 0; i-j>=0&&i+j+1<len;j++)  
            {  
                if(str[i-j]!=str[i+j+1])  
                    break;  
                if(2*j+2>maxlen)  
                    maxlen = 2*j+2;  
            }  
        }  
        printf("%d\n",maxlen);  
    }  
    return 0;  
}

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