I would like to evaluate a phone number using the provided method. The phone number should always have a length of 10. However the following method always seems to return false. Why is that? Thanks.
public static boolean valPhoneNumber(String phonenumber){
boolean result= true;
if (phonenumber.length() > 10 || phonenumber.length() < 10){
result= false;
}else
phonenumber.length();
char a=phonenumber.charAt(0);
char b=phonenumber.charAt(1);
char d=phonenumber.charAt(3);
char e=phonenumber.charAt(4);
char f=phonenumber.charAt(5);
if (a<2 || a>9){
result = false;
}else if( b<0 || b>8){
result = false;
}else if (d<2 || d>9){
result = false;
}else if (e==1 && f==1){
result = false;
}
return result;
}
So looking into your ladder which is comparing character to number. In this case the comparison will happen with ASCII value.
You can put single quotes to check the range:
if (a < '2' || a > '9') {
result = false;
} else if( b < '0' || b > '8') {
result = false;
} else if (d < '2' || d > '9') {
result = false;
} else if (e == '1' && f == '1') {
result = false;
}
One liner:
result = !((a < '2' || a > '9') || (b < '0' || b > '8') || (d < '2' || d > '9') || (e == '1' && f == '1'));
I think your code wrong at the parsing phonenumber.charAt(). This always return char, and when you do comparision with integer it will convert to number which present to that char code (ASCII code). I think you should modify your code to int a=Character.getNumericValue(phonenumber.charAt(0)); and so on
I think an approach with regex here would be the cleanest and easiest solution.
public static boolean valPhoneNumber(String phonenumber){
String regex = "[2-9][0-8][0-9][2-9][02-9][0-29][0-9]{4}";
return phonenumber.matches(regex);
}
You should cast the char variables to integer.
you can try this:
int a = Integer.parseInt(phonenumber.substring(0,1));
I added single quotes to check the range. Thank you all.
public static boolean valPhoneNumber(String phonenumber) {
boolean result= true;
if (phonenumber.length() != 10) {
result = false;
} else {
//phonenumber.length();
char a = phonenumber.charAt(0);
char b = phonenumber.charAt(1);
char d = phonenumber.charAt(3);
char e = phonenumber.charAt(4);
char f = phonenumber.charAt(5);
if (a < '2' || a > '9') {
} else if( b<'0' || b>'8') {
result = false;
} else if (d < '2' || d > '9') {
result = false;
} else if (e == '1' && f == '1') {
result = false;
}
}
return result;
}
Related
This question already has answers here:
What is a NumberFormatException and how can I fix it?
(9 answers)
Closed 2 years ago.
When I try to use the function to iterate the user input expression, I get the java.lang.NumberFormatException, I try fixing the loop much time, but I still cannot understand where did it when wrong. The IDE suggest it went wrong in the parstInt while loop
Here is the code:
import java.util.Scanner;
import java.util.Stack;
static Stack<Integer> stackForOperand = new Stack<Integer>();
static Stack<Character> stackForOperator = new Stack<Character>();
public static int processOneOperator(char stackForOperator, int num1, int num2) {
int result = 0;
switch (stackForOperator) {
case '+':
result = num1 + num2;
case '-':
result = num1 - num2;
case '*':
result = num1 * num2;
case '/':
if (num2 != 0) {
result = num1 / num2;
} else {
throw new UnsupportedOperationException("divide by zero error");
}
}
return result;
}
public static boolean num_order(char first, char second) {
if (first == '(' || second == ')') {
return false;
} else if ((first == '*' || first == '/') && (second == '+' || second == '-')) {
return false;
} else {
return true;
}
}
public static int calculation_loop(String expression) {
for (int i = 0; i < expression.length(); i++) {
if (expression.charAt(i) >= '0' && expression.charAt(i) <= '9') {
String more_num = "";
while (i < expression.length() && expression.charAt(i) >= '0' && expression.charAt(i) <= '9') {
more_num += expression.charAt(i++);
int more_num2 = Integer.parseInt(more_num);
stackForOperand.push(more_num2);
i--;
}
} else if (expression.charAt(i) == '(') {
stackForOperator.push(expression.charAt(i));
} else if (expression.charAt(i) == ')') {
while (stackForOperator.peek() != '(') {
stackForOperand.push(
processOneOperator(stackForOperator.pop(), stackForOperand.pop(), stackForOperand.pop()));
stackForOperator.pop();
}
} else if (expression.charAt(i) == '+' || expression.charAt(i) == '-' || expression.charAt(i) == '*'
|| expression.charAt(i) == '/') {
while (!stackForOperator.empty() && num_order(expression.charAt(i), stackForOperator.peek())) {
stackForOperand.push(
processOneOperator(stackForOperator.pop(), stackForOperand.pop(), stackForOperand.pop()));
stackForOperator.push(expression.charAt(i));
}
}
}
while (!stackForOperator.empty()) {
stackForOperand
.push(processOneOperator(stackForOperator.pop(), stackForOperand.pop(), stackForOperand.pop()));
}
return stackForOperand.pop();
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("/");
String input = scanner.nextLine();
input = input.replaceAll("\\s+", "");
System.out.println(input);
Integer output = calculation_loop(input);
System.out.println(output);
}
}
Looking at this piece of code:
public static int calculation_loop(String expression) {
for (int i = 0; i < expression.length(); i++) {
if (expression.charAt(i) >= '0' && expression.charAt(i) <= '9') {
String more_num = "";
while (i < expression.length() && expression.charAt(i) >= '0' && expression.charAt(i) <= '9') {
more_num += expression.charAt(i++);
int more_num2 = Integer.parseInt(more_num);
stackForOperand.push(more_num2);
i--;
}
So. Suppose having such an expression "2345+6789". According to your code i is incrementing and decrementing before end of while loop.
while (i < expression.length() && expression.charAt(i) >= '0' && expression.charAt(i) <= '9') {
more_num += expression.charAt(i++);
...
i--;
}
Is this your intention? It makes your loop infinite and it finishes only because number format exception is thrown while parsing. This is why I think your parser throws exception: You got the digit '2' in first place, concatenate it with more_num, then increment the i, after that decrement the i and at the next iteration you have the same previous position with same char '2', then concatenate it again to more_num ad infinitum. And first you parse more_num which is "2", on next iteration you append one more '2' and more_num "22" then "222"... Till it become bigger than type int can hold like "22222222222222222222222" and exception is thrown
Second thing. Suppose you remove i-- and your loop will normally iterate the next char. So, your stackForOperand will push first the number 2, then will push the number 23 then the number 234, then will push the number 2345. Is that your intension? I think more logically is to move Integer.parseInt and stackForOperand.push after the while loop
public static int calculation_loop(String expression) {
for (int i = 0; i < expression.length(); i++) {
if (expression.charAt(i) >= '0' && expression.charAt(i) <= '9') {
String more_num = "";
while (i < expression.length() && expression.charAt(i) >= '0' && expression.charAt(i) <= '9') {
more_num += expression.charAt(i++);
}
int more_num2 = Integer.parseInt(more_num);
stackForOperand.push(more_num2);
I tried to Evaluate Mathematical Expressions in Java with the following code:
public double applyOp(char op,double b,double a)
{
switch (op)
{
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
return a / b;
}
return 0;
}
public boolean hasPrecedence(char op1,char op2)
{
return (op1 != '*' && op1 != '/') || (op2 != '+' && op2 != '-');
}
public double evaluate(String input) {
Stack<Double> values = new Stack<>();
Stack<Character> ops = new Stack<>();
int stringIndex = 0;
while (stringIndex < input.length())
{
StringBuilder multiDigitsNumber = new StringBuilder();
// If the input is number put to stack values
if (input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
{
while (stringIndex < input.length() && input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
{
multiDigitsNumber.append(input.charAt(stringIndex++));
}
values.push(Double.parseDouble(multiDigitsNumber.toString()));
}
// If the input is operator put to stack ops
else
{
while (!ops.empty() && hasPrecedence(input.charAt(stringIndex),ops.peek()))
{
values.push(applyOp(ops.pop(),values.pop(),values.pop()));
}
ops.push(input.charAt(stringIndex++));
}
}
// Execute remain operator in stack values
while (!ops.empty()) {
values.push(applyOp(ops.pop(), values.pop(), values.pop()));
}
// The final number in stack value is result
return values.pop();
}
Input example:
12+24*2-30/5.....
The code above works fine but I wonder are there any way to replace
while (stringIndex < input.length() && input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
{
multiDigitsNumber.append(input.charAt(stringIndex++));
}
with something else so I don't have to use nested while loop in this situation. The goal is to catch number in string until it reach an operator
Thanks in advance
You can use Regex like this.
if (input.charAt(stringIndex) >= '0' && input.charAt(stringIndex) <= '9')
{
String number = input.substring(stringIndex).replaceAll("^(\\d+).*", "$1");
values.push(Double.parseDouble(number));
stringIndex += number.length();
}
I have no idea how to check if char[] contains only one letter (a or b) on the first position and only one int (0-8) on the second position. for example a2, b2
I have some this, but I do not know, what should be instead of digital +=1;
private boolean isStringValidFormat(String s) {
boolean ret = false;
if (s == null) return false;
int digitCounter = 0;
char first = s.charAt(0);
char second = s.charAt(1);
if (first == 'a' || first == 'b') {
if (second >= 0 && second <= '8') {
digitCounter +=1;
}
}
ret = digitCounter == 2; //only two position
return ret;
}
` public char[] readFormat() {
char[] ret = null;
while (ret == null) {
String s = this.readString();
if (isStringValidFormat(s)) {
ret = s.toCharArray();
}else {
System.out.println("Incorrect. Values must be between 'a0 - a8' and 'b0 - b8'");
}
}
return new char[0];
}`
First, I would test for null and that there are two characters in the String. Then you can use a simple boolean check to test if first is a or b and the second is between 0 and 8 inclusive. Like,
private boolean isStringValidFormat(String s) {
if (s == null || s.length() != 2) {
return false;
}
char first = s.charAt(0);
char second = s.charAt(1);
return (first == 'a' || first == 'b') && (second >= '0' && second <= '8');
}
For a well understood pattern, use Regex:
private static final Pattern pattern = Pattern.compile("^[ab][0-8]$")
public boolean isStringValidFormat(String input) {
if (input != null) {
return pattern.matcher(input).matches();
}
return false;
}
I need to find the number of distinct vowels. I came up with the code below but it can't make distinction between same vowels:
public static int count_Vowels(String str) {
str = str.toLowerCase();
int count = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == 'a' || str.charAt(i) == 'e' || str.charAt(i) == 'i'
|| str.charAt(i) == 'o' || str.charAt(i) == 'u') {
count++;
}
}
return count;
}
I would start with five variables (one for each vowel) set to 0, iterate the characters in the input and set the corresponding variable to 1 if I find a match, and simply return the accumulated value of said variables. Like,
public static int count_Vowels(String str) {
int a = 0, e = 0, i = 0, o = 0, u = 0;
for (char ch : str.toLowerCase().toCharArray()) {
if (ch == 'a') {
a = 1;
} else if (ch == 'e') {
e = 1;
} else if (ch == 'i') {
i = 1;
} else if (ch == 'o') {
o = 1;
} else if (ch == 'u') {
u = 1;
}
}
return a + e + i + o + u;
}
You could use Set data structure and instead of incrementing the counter just add vowels to the set. At the end you can return just the size of the set.
The problem in your code is that you are not counting the distinct vowels, but all the vowels in the string. A Java-8 way to this:
public static int countDistinctVowels(String str) {
str = str.toLowerCase();
int count = (int) str.chars() // get IntStream of chars
.mapToObj(c -> (char) c) // cast to char
.filter(c -> "aeiou".indexOf(c) > -1) // remove all non-vowels
.distinct() // keep the distinct values
.count(); // count the values
return count;
}
Also use proper Java naming conventions: countDistinctVowels, no count_Distinct_Vowels.
there's definitely an issue here with this counting. at the very least. you should rethink this:
if (str.charAt(i) == 'a' || str.charAt(i) == 'e' || str.charAt(i) == 'i'
|| str.charAt(i) == 'o' || str.charAt(i) == 'u')
count++;
You could use the method contains
public static int count_Vowels(String str) {
str = str.toLowerCase();
int count = 0;
count += string.contains("a") ? 1 : 0;
count += string.contains("e") ? 1 : 0;
count += string.contains("i") ? 1 : 0;
count += string.contains("o") ? 1 : 0;
count += string.contains("u") ? 1 : 0;
return count;
}
I made explanations in the comments to the code:
public static int count_Vowels(String str) {
str = str.toLowerCase();
Set<Character> setOfUsedChars = new HashSet<>(); // Here you store used vowels
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == 'a' || str.charAt(i) == 'e' || str.charAt(i) == 'i'
|| str.charAt(i) == 'o' || str.charAt(i) == 'u') { // if currently checked character is vowel...
setOfUsedChars.add(str.charAt(i)); // add this vowel to setOfUsedChars
}
}
return setOfUsedChars.size(); // size of this sets is a number of vowels present in input String
}
static void vow(String input){
String output=input.toLowerCase();
int flaga=0,flage=0,flagi=0,flago=0,flagu=0;
for(int i=0;i<input.length();i++) {
if((output.charAt(i))=='a' && flaga==0) {
System.out.print(input.charAt(i)+" ");
flaga++;
}
if(output.charAt(i)=='e' && flage==0) {
System.out.print(input.charAt(i)+" ");
flage++;
}
if(output.charAt(i)=='i' && flagi==0) {
System.out.print(input.charAt(i)+" ");
flagi++;
}
if(output.charAt(i)=='o' && flago==0) {
System.out.print(input.charAt(i)+" ");
flago++;
}
if(output.charAt(i)=='u' && flagu==0) {
System.out.print(input.charAt(i)+" ");
flagu++;
}
}
}
public static void main(String args[]) {
String sentence;
int v=0,c=0,ws=0;
Scanner sc= new Scanner(System.in);
sentence = sc.nextLine();
sc.close();
sentence.toLowerCase();
String res="";
for(int i=0;i<sentence.length();i++) {
if(sentence.charAt(i)=='a'||sentence.charAt(i)=='e'||sentence.charAt(i)=='i'||sentence.charAt(i)=='o'||sentence.charAt(i)=='u') {
if(res.indexOf(sentence.charAt(i))<0) {
res+=sentence.charAt(i);
v++;
}//System.out.println(res.indexOf(sentence.charAt(i)));
}
else if(sentence.charAt(i)==' ')
ws++;
else c++;
}
System.out.println(res);
System.out.println("no of vowels: "+v+"\n"+"no of consonants: "+c+"\n"+"no of
white spaces: "+ws);
}
You can use this Method to Find Count of Distinct vowels.
public static int count_Vowels(String str) {
char[] c = str.toLowerCase().toCharArray();
int Counter=0;
String NewString="";
for(int i=0;i<c.length;i++){
String tempString="";
tempString+=c[i];
if(!NewString.contains(tempString) && (c[i]=='a'||c[i]=='e'||c[i]=='i'||c[i]=='o'||c[i]=='u')){
Counter++;
NewString+=c[i];
}
}
return Counter;
}
Here is a solve for this problem without using objects. It's a crude but great solve for beginners who encounter this problem with limited js experience.
How to count unique vowels is a string;
function hasUniqueFourVowels(str){
let va = 0
let ve = 0
let vi = 0
let vo = 0
let vu = 0
let sum = 0
for(let i = 0; i < str.length; i++){
let char = str[i];
if(char === "i"){
vi = 1
}
if(char === "e"){
ve = 1
}
if(char === "a"){
va = 1
}
if(char === "o"){
vo = 1
}
if(char === "u"){
vu = 1
}
sum = va + vi + vo + ve + vu
if (sum >= 4){
return true
}
}
return false
}
#Test
public void numfindVoweles(){
String s="aeiouaedtesssiou";
char a[]=s.toCharArray();
HashMap<Character,Integer> hp= new HashMap<Character, Integer>();
for(char ch:a){
if(hp.containsKey(ch) && (ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u')){
hp.put(ch,hp.get(ch)+1);
}
else if(ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u'){
hp.put(ch,1);
}
}
System.out.println(hp);
}
This program is supposed to compare "DNA" strings.
Input:
3
ATGC
TACG
ATGC
CGTA
AGQ
TCF
First line represents how many times the program will be run. Each time it runs, it compares the two strings. A matches with T and vice versa. G matches with C and vise versa. So if the first letter of string 1 is A, the first letter of string 2 should be T. If the next one is T, the next one on the other string should be A and etc. If a letter other than A, T, G, or C appear, it is a bad sample. If its bad, print out bad, if its good, print out good. I tried many different combinations to this and they all worked fine but according the the judge's test data (they have different input), it failed. Does anyone see anything wrong with this? I know it might not be the most efficient way of getting the job done but it did, at least to my understanding.
Output:
GOOD
BAD
BAD
public class DNA
{
public static void main(String[] args) throws IOException
{
Scanner scan = new Scanner (new File ("dna.dat"));
int T = scan.nextInt();
scan.nextLine();
boolean valid = true;
for (int i = 0; i < T; i++)
{
String strand1 = scan.nextLine();
strand1 = strand1.toUpperCase();
String strand2 = scan.nextLine();
strand2 = strand2.toUpperCase();
for (int p = 0; p < strand1.length(); p++)
{
if (strand1.charAt(p) != 'A' && strand1.charAt(p) != 'T' && strand1.charAt(p) != 'G' && strand1.charAt(p) != 'C'
&& strand2.charAt(p) != 'A' && strand2.charAt(p) != 'T' && strand2.charAt(p) != 'G' && strand2.charAt(p) != 'C')
{
valid = false;
break;
}
if (strand1.length() != strand2.length())
{
valid = false;
break;
}
}
if (valid)
{
for (int p = 0; p < strand1.length(); p++)
{
if ((strand1.charAt(p) == 'A' && strand2.charAt(p) == 'T') || (strand1.charAt(p) == 'T' && strand2.charAt(p) == 'A')
|| (strand1.charAt(p) == 'G' && strand2.charAt(p) == 'C') || (strand1.charAt(p) == 'C' && strand2.charAt(p) == 'G'))
valid = true;
else
valid = false;
}
}
if (valid)
out.println("GOOD");
else
out.println("BAD");
valid = true;
}
}
}
I added the toUpperCase and compared the strings for equal length just as a last attempt to see if their data maybe had some lowercase letters or different length strings though they SHOULD all be the same length and uppercase. Nevertheless, the program was still rejected for "failing the judges test data."
You need a break in the second for loop when valid = false. For example if characters 1,2,3 are wrong but #4 is a match you will still end up with valid.
I would convert the strings to arrays to make things easier:
for (int i = 0; i < T; i++)
{
boolean valid = true;
String strand1 = scan.nextLine();
strand1 = strand1.toUpperCase();
String strand2 = scan.nextLine();
strand2 = strand2.toUpperCase();
if ( strand1.length() != strand2.length())
{
valid = false;
}
if (valid) {
char[] c1 = strand1.toCharArray();
char[] c2 = strand2.toCharArray();
for (int p = 0; p < c1.length; p++)
{
if (-1 == "ACTG".indexOf(c1[p]) || -1 == "ACTG".indexOf(c2[p]))
{
valid = false;
break;
}
}
if (valid)
{
for (int p = 0; p < c1.length; p++)
{
if (('A' == c1[p] && 'T' != c2[p]) ||
('T' == c1[p] && 'A' != c2[p]) ||
('C' == c1[p] && 'G' != c2[p]) ||
('G' == c1[p] && 'C' != c2[p])) {
valid = false;
break;
}
}
}
}
if (valid)
System.out.println("GOOD");
else
System.out.println("BAD");
}
Change all
&&
in
if (strand1.charAt(p) != 'A' && strand1.charAt(p) != 'T' && strand1.charAt(p) != 'G' && strand1.charAt(p) != 'C' && strand2.charAt(p) != 'A' && strand2.charAt(p) != 'T' && strand2.charAt(p) != 'G' && strand2.charAt(p) != 'C')
to
||
if ANY, not ALL character is other than A, T, G, or C, then we exit the loop.