【PAT】【Advanced Level】1040. Longest Symmetric String (25)

最新推荐文章于 2024-05-16 11:50:28 发布

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最新推荐文章于 2024-05-16 11:50:28 发布

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本文介绍了一种求解字符串中最长对称子串的方法，通过遍历字符串判断单数和双数情况来找到最长对称子串的长度。

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1040. Longest Symmetric String (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

原题链接：

https://www.patest.cn/contests/pat-a-practise/1040

思路：

循环一遍分别判断单数、双数的情况即可

CODE：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;

int main()
{
    string str;
    getline(cin,str);
    int ma=1;
    for (int i=0;i<str.length()-1;i++)
    {
        if (str[i]==str[i+1])
        {
            int t=0;
            while (i-t>=0&&i+t+1<str.length())
            {
                if (str[i-t]!=str[i+1+t]) break;
                t++;
            }
            ma=max(2*t,ma);
        }
        int tt=0;
        while (i-1-tt>=0 && i+tt+1<str.length())
        {
            if (str[i-1-tt]!=str[i+1+tt]) break;
            tt++;
        }
        ma=max(ma,1+tt*2);
    }
    cout<<ma<<endl;
    return 0;
}