Correct answer ,but why " no met !"

Question about the task

Shortest or longest

Java Syntax, Level 7, Lesson 6

Resolved

Don't understand why ? Please help ,thanks a lot!

1. Create a list of strings.
2. Add 10 strings from the keyboard.
3. Find out which string occurs earlier in the list: the shortest or the longest.
If several strings are shortest or longest, then consider the very first such string encountered.
4. Display the string described in Step 3. One string should be displayed.

Requirements:

Declare a string list variable and immediately initialize it.
The program should read 10 strings from the keyboard and add them to the list.
The program should display the shortest string if it comes before the longest.
The program should display the longest string if it comes before the shortest.

package com.codegym.task.task07.task0712; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; /* Shortest or longest */ public class Solution { public static void main(String[] args) throws Exception { BufferedReader r = new BufferedReader( new InputStreamReader(System.in)); ArrayList<String> list = new ArrayList<String>(10); for ( int i=0; i<10; i++) { String a= r.readLine(); list.add(a); } int ma = list.get(0).length(); for ( int i=0; i<10; i++) { if ( ma <list.get(i).length()) ma = list.get(i).length(); } int mi = list.get(0).length(); for ( int p=0; p<10; p++) { if ( mi > list.get(p).length()) mi = list.get(p).length(); } for (int i=0; i<10; i++) { if(ma ==list.get(i).length() || mi==list.get(i).length()) // System.out.println(list.get(i)); // else // if ( mi == list.get(i).length()) System.out.println(list.get(i)); break; } //write your code here } }

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Kamil

Level 31 , Near Warsaw, Poland

19 January 2020, 16:11

useful

Most simple way to solve this task is basically creating two variables which gonna contain index of longest and shortest string. And at the end of the program check which one is smaller and print it out. so just compare them in if statements like you would normally do and make sure you initalize these variables anytime condition is met.

Kamil

Level 31 , Near Warsaw, Poland

19 January 2020, 16:13

solution

int longest = listOfStrings.get(0).length();
        int shortest = listOfStrings.get(0).length();
        int indexShr = 0, indexLong = 0;

        for(int k = 0; k < listOfStrings.size(); k++) {
            if(listOfStrings.get(k).length() > longest) {
                longest = listOfStrings.get(k).length();
                indexLong = k;
            }
            if(listOfStrings.get(k).length() < shortest) {
                shortest = listOfStrings.get(k).length();
                indexShr = k;
            }
        }

Figure out next step.

Clarendon lee

Level 8 , Baltimore, United States

17 January 2020, 02:11

I think,this task makes some mistake! For example, I try several times, when I just use " if ( max== list.get(I).length())" it can output "grandfather " ! But when just use " if( min== list.get(I).length())" it can't output anything at all ! This task is wrong at all. ,!!!

jajaceek

Level 23 , Warszawa, Poland

12 January 2020, 07:39

useful

try this way: for(int j = 0; j < list.size(); j++){ if(list.get(j).length() == ma || list.get(j).length() == mi) { System.out.println(list.get(j)); break; } also in line 22 and 27 it should be: int i = 0

Clarendon lee

Level 8 , Baltimore, United States

12 January 2020, 23:29

Thanks for your help ! I changed my code that according to what you said. But test is still " is no met " !

jajaceek

Level 23 , Warszawa, Poland

13 January 2020, 04:15

try this: ArrayList<String> list = new ArrayList<>(); int max = 0; int min = 100; BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); for (int i = 0; i < 10; i++){ String text = reader.readLine(); list.add(text); int d = text.length(); if (max < d) max = d; if (min > d) min = d;