2017年上海金马五校程序设计竞赛：Problem E : Find Palindrome （字符串处理）

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

分析：

题意：给出一个字符串，求这个字符串中的最长回文串的长度。

首先要一个一个看字符串中的字符：

对于第 i 个字符 str[i] ，假如回文子串是奇数个字符，那么考虑以i为中心，同时向左向右扩展，直到发现对称位置字符不相等，假如此时共扫过x个字符，则当前回文串长度为2*x+1。加上的1就是加上i本身。如下图所示：

由第 i 个字符a[i]构成回文偶数串。i在最接近对称轴的左侧。

代码：

#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
",maxlen);
    }
    return 0;
}