找出两个字符串的最长公共前缀

Question

I want to find the longest common prefix of two strings. Is there a way to loop my last couple of if statements so that I can end at the last characters that do not match each other?

System.out.println("Enter the first string: ");
String s = input.nextLine();

System.out.println("Enter the second string: ");
String s2 = input.nextLine();

//check if first characters are same
if (s.charAt(0) != s2.charAt(0)) {
  System.out.println(""+s+ " and "+s2+ " have no common prefix");
  System.exit(0);
    }

if (s.charAt(0) == s2.charAt(0))
  System.out.print(" "+s.charAt(0));

if (s.charAt(0) == s2.charAt(0))
  System.out.print(" "+s.charAt(1));

if (s.charAt(0) == s2.charAt(0))
  System.out.print(" "+s.charAt(2));  
  }
}

Example:

Enter first string: Welcome to c++

Enter second string: Welcome to java

The code should return Welcome to as the common prefix.

Answer 1

Maybe something like:

int sLength = s.length(),
    s2Length = s2.length(),
    minLength = (sLength < s2Length) ? sLength : s2Length;

for (int i = 0; i < minLength; i++) {
    if (s.charAt(i) == s2.charAt(i)) {
        System.out.println(s.charAt(i));
    }
    else {
        break;
    }
}

But more details about your question would be great.

Edit: It depends what @afrojuju_ wants to do. That's not clear. Some more logic may be added to accomplish the desired behavior.

Edit 2: Added string length comparison as pointed out by @JavaBeast.

Answer 2

try this. I guess this is what you are trying to achieve. If this is correct I will add explanation later

import java.util.*;
import java.lang.*;
import java.io.*;

class Ideone
{
    public static void main (String[] args) throws java.lang.Exception
    {
        String s = "Hello Wo";
        String s2 = "Hello World";
        String small,large;
         if(s.length() > s2.length()) 
            {small = s2;large = s;}
          else
            {small = s;large = s2;}
        int index = 0;    
        for(char c: large.toCharArray())
        {
            if(index==small.length()) break;
            if(c != small.charAt(index)) break;
            index++;
        }
        if(index==0)
          System.out.println(""+s+ " and "+s2+ " have no common prefix");
        else
          System.out.println(large.substring(0,index));
    }
}

Edit:

1. I find the larger of the strings and choose it to be the outer string to loop throught
2. toCharArray() converts the string into characters so you can loop through each characters in the string using Java's foreach (For more click [1] )
3. Inside the loop you should exit on two conditions
   • End of the string (I use length to find if I reach end of smaller string)
   • no more matching characters between two strings
4. you increment the index until you break out in one of the above conditions
5. By the time you break out of the for loop, index will contain the last index where both string are continuously equal.
6. If index = 0. just say no matches else print characters from 0 until index

Answer 3

    public static String LcpFinder (String s1 , String s2){

    if (s1 == null || s2 == null){
        throw new IllegalArgumentException();
    }

    int minLength = 0;

    if (s1.length() < s2.length()){
        minLength = s1.length();
    }
    else{
        minLength = s2.length();
    }

    for (int i = 0 ; i < minLength ; i++){

        if(s1.charAt(i) == s2.charAt(i)){
            continue;
        }
        else{
            return s1.substring(0,i);
        }           
    }
    return s1.substring(0,minLength); 
}

Answer 4

str1 = input().lower() str2 = input().lower()

for i in range(min(len(str1),len(str2))): if str1[i]:= str2[i]: break

if i == 0: print(-2) else: print(str1[:i])