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
题意:给定一个字符串,让你找出其中最长的对称子串。
思路1:字符串中可能包含空格因此要使用getline(cin,string),定义一个判断最长对称长度的函数,每次根据当前下标向两边拓展长度,分别进行奇对称和偶对称判断。最后输出最大值即可。初始下标从1开始。
参考代码:
#include<algorithm>
#include<iostream>
#include<string>
using namespace std;
int getMaxSymLength(string s,int k){
int oddLength=1,evenLength=0;
int i=k-1,j=k+1;
while(i>=0&&j<s.size()&&s[i]==s[j]){
i--;j++;
oddLength+=2;
}
i=k-1,j=k;
while(i>=0&&j<s.size()&&s[i]==s[j]){
i--;j++;
evenLength+=2;
}
return max(oddLength,evenLength);
}
int main()
{
int maxLength=0,temp=0;
string s;
getline(cin,s);
if(s.size()<2) cout<<s.size()<<endl;
else{
for(int i=1;i<s.size();i++)
{
temp=getMaxSymLength(s,i);
maxLength=max(maxLength,temp);
}
cout<<maxLength<<endl;
}
return 0;
}
思路2:动态规划解法,设数组dp[i][j]表示s[i]到s[j]是否是对称串。其数值只能是0或1.状态转移方程如下:
1.dp[i][j]=dp[i+1][j-1] 若s[i]==s[j]
2.dp[i][j]=0 若s[i]!=s[j]
初始化:dp[i][i]=1,即长度为1的字串一定对称。如果s[i]==s[i+1],dp[i][i+1]=1,,将长度为2的对称串标出。
然后按照字串长度L=3直到L=s.size(),分别进行迭代计算,这样当求较长的对称子串长度时,则可以利用较短的对称子串结果。例如长度为3的dp值求出后,再求长度为4的dp值时,则可以直接根据s[i]==s[j]是否成立,进行状态转移。
参考代码:
#include<string>
#include<iostream>
using namespace std;
const int maxn=1010;
int dp[maxn][maxn]={0};
int main()
{
int maxLength=0;
string s;
getline(cin,s);
for(int i=0;i<s.size();i++){
dp[i][i]=1;
maxLength=1;
if(i<s.size()-1&&s[i]==s[i+1]){
dp[i][i+1]=1;
maxLength=2;
}
}
for(int l=3;l<=s.size();l++)
for(int i=0;i<=s.size()-l;i++){
int j=i+l-1;
if(s[i]==s[j]&&dp[i+1][j-1]==1){
dp[i][j]=1;
maxLength=l;
}
}
cout<<maxLength<<endl;
return 0;
}