1040 Longest Symmetric String (25 point(s))

1040 Longest Symmetric String (25 point(s))

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

 最长回文子串

方法一：暴力求解

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main(void){
	string s;getline(cin,s);
	int maxLen = 1; 
	for(int i=0;i<s.length();i++){
		for(int j=i+1;j<=s.length();j++){
			if(j-i>maxLen){
				string sub = s.substr(i,j-i);
				string origin = sub;
				reverse(sub.begin(),sub.end()); 
				if(origin == sub){
					maxLen = j-i;
				}
			}
		}
	}
	cout<<maxLen<<endl;
	return 0;
}

 方法2：遍历字符串，向两边扩展（注意分奇数长度、偶数长度两种情况）

参考链接：https://blog.youkuaiyun.com/richenyunqi/article/details/80218079

#include<bits/stdc++.h>
using namespace std;
int main(){
    string s;
    getline(cin,s);
    int maxLen=0;
    for(int i=0;i<s.size();++i){
        int j;
        for(j=1;i-j>=0&&i+j<s.size()&&s[i+j]==s[i-j];++j);//以当前字符为回文中心查找最长回文子串
        maxLen=max(maxLen,2*j-1);//更新回文子串最大长度
        for(j=0;i-j>=0&&i+j+1<s.size()&&s[i-j]==s[i+1+j];++j);//以当前字符为回文中心左侧字符查找最长回文子串
        maxLen=max(maxLen,2*j);//更新回文子串最大长度
    }
    printf("%d",maxLen);
    return 0;
}