Topic description:
Given n pairs of strings, find the greatest common substring in each pair.
Input format:
2 //Indicates that there are 2 groups of
fsjdfgjs //The first group of
fdfg
jfdslkdfj //The second group
fjdkdfs
Output format:
3
3
The problem is very simple to understand, but it is still very complicated to solve the problem in an optimal way. I am afraid that the KMP algorithm will come to mind when you see the common substring. I saw in "The Optimal Solution to the Problem of Algorithms and Data Structures of IT Famous Enterprises" that the time complexity of the solution is O(M*N) and the additional space complexity is O(1).
However, due to the limited time for online programming, the method I implemented is very simple, that is, 1. Find the shorter string, and then take the substrings from it in turn. This is a bit tricky to take the substrings in turn. The longest one should be taken first and then gradually shortened. , which makes it the longest common substring if found; 2. Compare the substring with the longer string in turn, and if it contains, it is the longest common substring
code show as below:
import java.util.Scanner;
public class SecondQuestion {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String number = in.nextLine();
int num = Integer.parseInt(number);
String[] str = new String[num*2];
for(int i=0; i<num*2; i++){
str[i] = in.nextLine();
str[++i] = in.nextLine();
}
String[] arr = new String[num];
int j =0;
for(int i=0; i<num*2; i++){
arr[j] = Max(str[i], str[++i]);
if(arr[j] == null){
System.out.println(0);
}else{
System.out.println(arr[j].length());
}
j++;
}
}
// the largest common substring of the two strings
public static String Max(String s1, String s2) {
String max = (s1.length() > s2.length()) ? s1 : s2;
String min = max.equals(s1) ? s2 : s1;
for (int i = 0; i < min.length(); i++) {
for (int m=0,n=min.length()-i ; n!=min.length()+1; m++, n++) {
String sub = min.substring(m, n);
if (max.contains(sub)) {
return sub;
}
}
}
return null;
}
}
Result screenshot:
