Programming questions

Violence enumeration  

• We find each seed sequence from a shorter string each time （ Successive ）, To compare in a longer string , Maintain the longest length every time , If it is greater than the longest length , Just compare this string
• Need to distinguish between long and short strings

import java.util.Scanner;
public class Main{
      public static void main(String[] args) {
        Scanner sc =new Scanner(System.in);
        String str1=sc.nextLine();
        String str2=sc.nextLine();
        String str11=str1.length()<str2.length()?str1:str2;
        String str22=str1.length()<str2.length()?str2:str1;
        //str1 Is a shorter string  str2 Is a long string 
        int max=0;
        for (int i = 0; i <str11.length(); i++) {
            for (int j = i+1; j <=str11.length() ; j++) {
                if (str22.contains(str11.substring(i,j))&&j-i>max){
                    max=j-i;
                }
            }
        }
        System.out.println(max);
    }
}

Dynamic programming

import java.util.Scanner;
public class Main{
        public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        String str1=sc.nextLine();
        String str2=sc.nextLine();
        int row=str1.length();
        int column=str2.length();
        int dp[][]=new int[row+1][column+1];
        int max=0;
        //dp[i][j] Express str1 Before i Two strings and str2 Before j Maximum common string length of strings 
        //base case
        for (int i = 0; i <=row; i++) {
            dp[i][0]=0;
        }
        for (int i = 0; i <=column; i++) {
            dp[0][i]=0;
        }
        for (int i = 1; i <=row; i++) {
            for (int j = 1; j <=column ; j++) {
                if(str1.charAt(i-1)!=str2.charAt(j-1)){
                    dp[i][j]=0;
                }else {
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                max=Math.max(max,dp[i][j]);
            }
        }
        System.out.println(max);
    }
}