有模式 *A*A'* ,其中*表示任意子字符串(可以是空串),A也表示任意子字符串(也可以是空串),A'是A的逆序,例如假设A是23ad4那么A'就是4da32。
这样,任意字符串都符合这个模式,而且会有多个匹配方法,例如字符串
13412432145
可以匹配为 1 34 1 2432145 其中A=1
也可以匹配为 1 34 12 43 2145 其中A=34
也可以匹配为 13 412 43 214 5 其中A=412
问题:写一个程序,输入任意字符串X=*A*A'*,求长度最大的A
例如输入 13412432145 则最大长度A=412,长度等于3
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using System;
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using System.Text;
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using System.IO;
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class Class1
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{
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/// <summary>
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/// 应用程序的主入口点。
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/// </summary>
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[STAThread]
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static void Main(string[] args)
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{
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int count=0;
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Console.WriteLine("Input the test string:");
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string s = Console.ReadLine();
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bool found = false;
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while(s!=null && s!="")
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{
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int n = s.Length;
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for(int m=n/2;m>0;m--)
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{
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for(int i=0;i<n-m;i++)
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{
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string a = s.Substring(i,m);
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string a_ = Reverse(a);
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int index = s.IndexOf(a_,i+m);
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count++;
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if(index!=-1)
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{
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Console.WriteLine("比较次数:"+count);
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Console.WriteLine("A= "+a);
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Console.WriteLine("模式为");
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if(i-1>0)
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Console.Write(s.Substring(0,i-1)+" ");
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Console.Write(a+" ");
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if(index-i-m>0)
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Console.Write(s.Substring(i+m,index-i-m)+ " ");
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Console.Write(a_+ " ");
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Console.WriteLine(s.Substring(index+m));
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found = true;
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goto label;
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}
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}
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}
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label:
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if(found == false)
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Console.WriteLine(@"a=""");
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Console.WriteLine("Inpu the test string:");
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s = Console.ReadLine();
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}
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}
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static string Reverse(string s)
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{
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StringBuilder sb = new StringBuilder();
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for(int i=s.Length-1;i>=0;i--)
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{
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sb.Append(s[i]);
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}
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return sb.ToString();
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}
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}
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