scanner.nextInt() returns java.util.NoSuchElementException - java.util.scanner

This is my code, which is supposed to accept input from the user and set 2 int values. The exit function works correctly however when the input string is "5 2" for example, it sets x as 5 and throws the java.util.NoSuchElementException at the y = s2.nextInt(); line, even though there is a next int. In an example of nextInt() I saw, ints were seperated by a space and the scanner still picked up all the integers. Is mine missing something?
String exit = "-1";
Scanner s1 = new Scanner(System.in);
String input = s1.next();
Scanner s2 = new Scanner(input);
if (input.equals(exit))
Sequence.quit();
else {
x = s2.nextInt();
y = s2.nextInt();
}

If you print out your variable ìnput, you will see that it contains only "5" - since s1.next() wil stop at space.
Something like this:
String exit = "-1";
Scanner s1 = new Scanner(System.in);
int x = s1.nextInt();
if (x==-1)
Sequence.quit();
else {
y = s1.nextInt();
}

Related

how to extract an integer range from a string

I have a string that contains different ranges and I need to find their value
var str = "some text x = 1..14, y = 2..4 some text"
I used the substringBefore() and substringAfter() methodes to get the x and y but I can't find a way to get the values because the numbers could be one or two digits or even negative numbers.
One approach is to use a regex, e.g.:
val str = "some text x = 1..14, y = 2..4 some text"
val match = Regex("x = (-?\\d+[.][.]-?\\d+).* y = (-?\\d+[.][.]-?\\d+)")
.find(str)
if (match != null)
println("x=${match.groupValues[1]}, y=${match.groupValues[2]}")
// prints: x=1..14, y=2..4
\\d matches a single digit, so \\d+ matches one or more digits; -? matches an optional minus sign; [.] matches a dot; and (…) marks a group that you can then retrieve from the groupValues property. (groupValues[0] is the whole match, so the individual values start from index 1.)
You could easily add extra parens to pull out each number separately, instead of whole ranges.
(You may or may not find this as readable or maintainable as string-manipulation approaches…)
Is this solution fit for you?
val str = "some text x = 1..14, y = 2..4 some text"
val result = str.replace(",", "").split(" ")
var x = ""; var y = ""
for (i in 0..result.count()-1) {
if (result[i] == "x") {
x = result[i+2]
} else if (result[i] == "y") {
y = result[i+2]
}
}
println(x)
println(y)
Using KotlinSpirit library
val rangeParser = object : Grammar<IntRange>() {
private var first: Int = -1
private var last: Int = -1
override val result: IntRange
get() = first..last
override fun defineRule(): Rule<*> {
return int {
first = it
} + ".." + int {
last = it
}
}
}.toRule().compile()
val str = "some text x = 1..14, y = 2..4 some text"
val ranges = rangeParser.findAll(str)
https://github.com/tiksem/KotlinSpirit

Program to find if a string is a palindrome keeps on failing. Even after using toLowerCase() command for both strings, output doesn't come

import java.util.Scanner;
class Palindrome_string
{
public static void main()
{
System.out.println("\f");
Scanner sc = new Scanner(System.in);
System.out.println("Enter a string");
String a = sc.nextLine();
int b = a.length();
String rev = "";
for (int i = b - 1; i >= 0; i--)
{
char c = a.charAt(i);
rev = rev + c;
}
System.out.println("Original word "+a);
System.out.println("Reversed word "+rev);
a = a.toLowerCase();
rev = rev.toLowerCase();
if (a == rev)
{
System.out.println("It is a palindrome");
}
else
{
System.out.println("It is not a palindrome");
}
sc.close();
}
}
The program compiles properly. Still, when running the program, the message which tells if it is a palindrome prints incorrectly. What changes do I make? Here is a picture of the output. Even though the word 'level' (which is a palindrome) has been inputted, it shows that it isn't a palindrome. What changes should I make? output pic
You should not use == to compare two strings because it compares the reference of the string, i.e. whether they are the same object or not.
Use .equals() instead. It tests for value equality. So in your case:
if (a.equals(rev))
{
System.out.println("It is a palindrome");
}
Also try not to use single-letter variable names except for index variables when iterating over a list etc. It's bad practice.

Error in using indexOf not finding char in Arduino String

I have some code that I have no clue why it isn't working.
The code takes a serial input in the form of "xxx,yyy,zzz", where digits can range from 1 to 3 in each number. Because of an odd quirk in an app, it needs to be read as a char, then converted to a string to be handled. The intention is to split into 3 ints, red green and blue, from "RRR,GGG,BBB".
Now this works fine when I manually define String str (see commented code), but when I go and enter it from the serial console, it doesn't want to work. It seems to be coming from the indexOf(',') part, as while using Serial.print(c1);, I found that when I manually entered a string, it returned an index of the comma, but when I used the serial console, it returned -1 (not found).
And yes, the entered string into the console is in the correct format of "RRR,GGG,BBB", I've confirmed that by printing both phone and str independently.
while (Serial.available() > 0) {
char phone = Serial.read();
String str = String(phone);
//String str = "87,189,183";
int ln = str.length()-1;
int c1 = str.indexOf(','); //first place to cut string
int c2 = str.indexOf(',',c1+1); //second place
red = str.substring(0,c1).toInt();
green = str.substring(c1,c2).toInt();
blue = str.substring(c2,ln).toInt();
Serial.print(red);
Edit: With the Arduino String class, creating a string from a char is returning more than just one character, eleven in fact.
This:
char phone = Serial.read();
String str = String(phone);
will never create a string in str that has more than 1 character, since that's what you say you want.
This is the code for the Arduino's String(char) constructor:
String::String(char c)
{
init();
char buf[2];
buf[0] = c;
buf[1] = 0;
*this = buf;
}
So clearly your code will create a 1-character long string.
Also, beware of using indexes computed on the full string, on substrings later.
I'm try to guess that you are using these serial API http://playground.arduino.cc/Interfacing/CPPWindows.
while (Serial.available() > 0) {
char buf[12];
int len = Serial.ReadData(buf,11);
String str = String(buf);
//String str = "87,189,183";
int ln = str.length()-1;
int c1 = str.indexOf(','); //first place to cut string
int c2 = str.indexOf(',',c1+1); //second place
red = str.substring(0,c1).toInt();
green = str.substring(c1,c2).toInt();
blue = str.substring(c2,ln).toInt();
Serial.print(red);
If you are using other API like http://arduino.cc/en/Serial/Read you should follow these API where Serial is a Stream and read() return just the first available char.
Code was fixed by using a different function.
while (Serial.available() > 0) {
char phone = Serial.read();
str += phone;
//String str = "87,189,183";
int ln = str.length()-1;
int c1 = str.indexOf(','); //first place to cut string
int c2 = str.indexOf(',',c1+1); //second place
red = str.substring(0,c1).toInt();
green = str.substring(c1,c2).toInt();
blue = str.substring(c2,ln).toInt();
Serial.print(red);
I'm not sure why this works, and why before I was getting a string with more than one character. But it works!

How to split a string into multiple strings if spaces are detected (GM:Studio)

I made a console program, but the problem is that it doesn't allow parameters to be inserted. So I'm wondering how would I split a single string into multiple strings to achieve what I need. E.g.: text="msg Hello" would be split into textA="msg" and textB="Hello"
This is the main console code so far (just to show the idea):
if (keyboard_check_pressed(vk_enter)) {
text_console_c = asset_get_index("scr_local_"+string(keyboard_string));
if (text_console_c > -1) {
text_console+= "> "+keyboard_string+"#";
script_execute(text_console_c);
text_console_c = -1;
}
else if (keyboard_string = "") {
text_console+= ">#";
}
else {
text_console+= "> Unknown command: "+keyboard_string+"#";
};
keyboard_string = "";
}
I cant recommend spliting string with iteration by char, because when u try split very very very long string, then time to split is very long and can freeze thread for a short/long time. Game maker is single threaded for now.
This code is much faster.
string_split
var str = argument[0] //string to split
var delimiter = argument[1] // delimiter
var letDelimiter = false // append delimiter to each part
if(argument_count == 3)
letDelimiter = argument[2]
var list = ds_list_create()
var d_at = string_pos(delimiter, str)
while(d_at > 0) {
var part = string_delete(str, d_at , string_length(str))
if(letDelimiter)
part = part + delimiter
str = string_delete(str, 1, d_at)
d_at = string_pos(delimiter, str)
ds_list_add(list, part)
if(d_at == 0 && str != "")//last string without delimiter, need to add too
ds_list_add(list, str)
}
return list;
Dont forget ds_list_destroy after you iterate all strings
for example:
var splited = string_split("first part|second part", '|')
for(splited) {
//do something with each string
}
ds_list_destroy(splited)
Something like this may help, haven't tested it out but if you can follow what is going on its a good place to start.
Text = "msg Hello"
counter = 0
stringIndex = 0
for (i = 0; i < string_length(text); i++)
{
if string_char_at(text,i) == " "
{
counter++
stringIndex = 0
} else {
string_insert(string_char_at(text,i),allStrings(counter),stringIndex)
stringIndex++
}
}
allStrings should be an array containing each of the separate strings. Whenever a " " is seen the next index of allStrings starts having it's characters filled in. stringIndex is used to add the progressive characters.

Finding minimum moves required for making 2 strings equal

This is a question from one of the online coding challenge (which has completed).
I just need some logic for this as to how to approach.
Problem Statement:
We have two strings A and B with the same super set of characters. We need to change these strings to obtain two equal strings. In each move we can perform one of the following operations:
1. swap two consecutive characters of a string
2. swap the first and the last characters of a string
A move can be performed on either string.
What is the minimum number of moves that we need in order to obtain two equal strings?
Input Format and Constraints:
The first and the second line of the input contains two strings A and B. It is guaranteed that the superset their characters are equal.
1 <= length(A) = length(B) <= 2000
All the input characters are between 'a' and 'z'
Output Format:
Print the minimum number of moves to the only line of the output
Sample input:
aab
baa
Sample output:
1
Explanation:
Swap the first and last character of the string aab to convert it to baa. The two strings are now equal.
EDIT : Here is my first try, but I'm getting wrong output. Can someone guide me what is wrong in my approach.
int minStringMoves(char* a, char* b) {
int length, pos, i, j, moves=0;
char *ptr;
length = strlen(a);
for(i=0;i<length;i++) {
// Find the first occurrence of b[i] in a
ptr = strchr(a,b[i]);
pos = ptr - a;
// If its the last element, swap with the first
if(i==0 && pos == length-1) {
swap(&a[0], &a[length-1]);
moves++;
}
// Else swap from current index till pos
else {
for(j=pos;j>i;j--) {
swap(&a[j],&a[j-1]);
moves++;
}
}
// If equal, break
if(strcmp(a,b) == 0)
break;
}
return moves;
}
Take a look at this example:
aaaaaaaaab
abaaaaaaaa
Your solution: 8
aaaaaaaaab -> aaaaaaaaba -> aaaaaaabaa -> aaaaaabaaa -> aaaaabaaaa ->
aaaabaaaaa -> aaabaaaaaa -> aabaaaaaaa -> abaaaaaaaa
Proper solution: 2
aaaaaaaaab -> baaaaaaaaa -> abaaaaaaaa
You should check if swapping in the other direction would give you better result.
But sometimes you will also ruin the previous part of the string. eg:
caaaaaaaab
cbaaaaaaaa
caaaaaaaab -> baaaaaaaac -> abaaaaaaac
You need another swap here to put back the 'c' to the first place.
The proper algorithm is probably even more complex, but you can see now what's wrong in your solution.
The A* algorithm might work for this problem.
The initial node will be the original string.
The goal node will be the target string.
Each child of a node will be all possible transformations of that string.
The current cost g(x) is simply the number of transformations thus far.
The heuristic h(x) is half the number of characters in the wrong position.
Since h(x) is admissible (because a single transformation can't put more than 2 characters in their correct positions), the path to the target string will give the least number of transformations possible.
However, an elementary implementation will likely be too slow. Calculating all possible transformations of a string would be rather expensive.
Note that there's a lot of similarity between a node's siblings (its parent's children) and its children. So you may be able to just calculate all transformations of the original string and, from there, simply copy and recalculate data involving changed characters.
You can use dynamic programming. Go over all swap possibilities while storing all the intermediate results along with the minimal number of steps that took you to get there. Actually, you are going to calculate the minimum number of steps for every possible target string that can be obtained by applying given rules for a number times. Once you calculate it all, you can print the minimum number of steps, which is needed to take you to the target string. Here's the sample code in JavaScript, and its usage for "aab" and "baa" examples:
function swap(str, i, j) {
var s = str.split("");
s[i] = str[j];
s[j] = str[i];
return s.join("");
}
function calcMinimumSteps(current, stepsCount)
{
if (typeof(memory[current]) !== "undefined") {
if (memory[current] > stepsCount) {
memory[current] = stepsCount;
} else if (memory[current] < stepsCount) {
stepsCount = memory[current];
}
} else {
memory[current] = stepsCount;
calcMinimumSteps(swap(current, 0, current.length-1), stepsCount+1);
for (var i = 0; i < current.length - 1; ++i) {
calcMinimumSteps(swap(current, i, i + 1), stepsCount+1);
}
}
}
var memory = {};
calcMinimumSteps("aab", 0);
alert("Minimum steps count: " + memory["baa"]);
Here is the ruby logic for this problem, copy this code in to rb file and execute.
str1 = "education" #Sample first string
str2 = "cnatdeiou" #Sample second string
moves_count = 0
no_swap = 0
count = str1.length - 1
def ends_swap(str1,str2)
str2 = swap_strings(str2,str2.length-1,0)
return str2
end
def swap_strings(str2,cp,np)
current_string = str2[cp]
new_string = str2[np]
str2[cp] = new_string
str2[np] = current_string
return str2
end
def consecutive_swap(str,current_position, target_position)
counter=0
diff = current_position > target_position ? -1 : 1
while current_position!=target_position
new_position = current_position + diff
str = swap_strings(str,current_position,new_position)
# p "-------"
# p "CP: #{current_position} NP: #{new_position} TP: #{target_position} String: #{str}"
current_position+=diff
counter+=1
end
return counter,str
end
while(str1 != str2 && count!=0)
counter = 1
if str1[-1]==str2[0]
# p "cross match"
str2 = ends_swap(str1,str2)
else
# p "No match for #{str2}-- Count: #{count}, TC: #{str1[count]}, CP: #{str2.index(str1[count])}"
str = str2[0..count]
cp = str.rindex(str1[count])
tp = count
counter, str2 = consecutive_swap(str2,cp,tp)
count-=1
end
moves_count+=counter
# p "Step: #{moves_count}"
# p str2
end
p "Total moves: #{moves_count}"
Please feel free to suggest any improvements in this code.
Try this code. Hope this will help you.
public class TwoStringIdentical {
static int lcs(String str1, String str2, int m, int n) {
int L[][] = new int[m + 1][n + 1];
int i, j;
for (i = 0; i <= m; i++) {
for (j = 0; j <= n; j++) {
if (i == 0 || j == 0)
L[i][j] = 0;
else if (str1.charAt(i - 1) == str2.charAt(j - 1))
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = Math.max(L[i - 1][j], L[i][j - 1]);
}
}
return L[m][n];
}
static void printMinTransformation(String str1, String str2) {
int m = str1.length();
int n = str2.length();
int len = lcs(str1, str2, m, n);
System.out.println((m - len)+(n - len));
}
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String str1 = scan.nextLine();
String str2 = scan.nextLine();
printMinTransformation("asdfg", "sdfg");
}
}

Resources