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My code is:
int diff = 0;
for (int i = 0; i<listOne.size(); i++)
{
for (int j = 0; j<listTwo.size(); j++)
{
if (listOne.get(i)-listTwo.get(j)>diff)
diff = listOne.get(i)-listTwo.get(j);
if (listTwo.get(j)-listOne.get(i)>diff)
diff = listTwo.get(j)-listOne.get(i);
}
}
return diff;
The task is to find the greatest difference between any two numbers in two inputted lists (the difference must be between a number from list one and a number from list two).
I cannot tell what is wrong with my code.
You may be missing Math.abs executed on the diff. Difference is an absolute value, so difference between 5 to 7 and 7 to 5 is same - 2.
int diff = 0;
for (int i = 0; i<listOne.size(); i++) {
for (int j = 0; j<listTwo.size(); j++) {
int elementDiff = Math.abs(listOne.get(i)-listTwo.get(j));
if (elementDiff>diff) {
diff = elementDiff;
}
}
}
return diff;
So both below lines will produce same results:
int elementDiff = Math.abs(listOne.get(i)-listTwo.get(j));
int elementDiff = Math.abs(listTwo.get(i)-listOne.get(j));
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I need to sort by cakes made then if if equal by alphabetical order
I need this sorting method to firstly sort my cakesMade which is an int and then secondly by name if cakesmade is equal. The code is working for the cakesMade, I just cant seem to get the second sort by name to work. Could someone please help
public int bubbleSort (){
for (int i = 0; i < team.size(); i++){
for (int j = 0; j < team.size() - 1;j++ ){
Employee t1 = team.get(j);
Employee t2 = team.get(j+1);
if (t1.getCakesMade() < t2.getCakesMade()) {
team.set(j, t2);
team.set(j + 1, t1);
} else if (t1.getCakesMade() == t2.getCakesMade()) {
if (t1.getName().compareTo(t2.getName()) < 0){
team.set(j, t2);
team.set(j + 1, t1);
}
}
}
}
return 0;
}
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I am using java
For example, the following situation:
First, the function is used as a print all possible strings.
char[] alphabetSet = "012abc".toCharArray();
int length =5;
Output:
0,1,2,a,b,c,01,02,0a,0b,0c,10,11,12,1a ..................... ccccc. stop in length = 5
Then, I want to add a loop stopper to fetch the specified string.
char[] alphabetSet = "012abc".toCharArray();
int length =5;
int loopStopper = 3;
Output:
a
Thank you
Use backtracking.
void print_all(char []ch,int maxLen){
for(int i=1;i<=maxLen;i++)
backTrack(ch,i,0,new char[i]);
}
void backTrack(char[] ch,int len,int k,char[] ans){
if(k==len){
System.out.print(new String(ans,0,len)+",");
return;
}
for(int i=0;i<ch.length;i++){
ans[k]=ch[i];
backTrack(ch,len,k+1,ans);
}
}
try this:
String alphabet = "012abc";// for example as your code "012abc"
char[] alphabetSet = alphabet.toCharArray();
int length = 5;
for (int i = 0; i < alphabetSet.length; i++) {
System.out.print(alphabetSet[i] + ",");
}
for (int j = 0; j <= length; j++) {
for (int i = 0; i < alphabetSet.length; i++) {
System.out.printf("%d%c,",j,alphabetSet[i]);
}
}
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so my real question is, how can i make this code identify all the "look alike" numbers while theire running from 1 to 99, for example :11,22,33,44,...
and while the program identify them it sends a message.
package doodle;
int num2=11;
for (int i=1; i<100; i++) {
System.out.println(i);
int num1=i;
if(num1==num2) {
System.out.println("WOW");
}
}
Thanks
I would do using a String
for (int i = 11; i < 100; i++) {
StringBuffer orig = new StringBuffer();
String left = orig.append(i).toString();
if (orig.reverse().toString().equals(left)) {
System.out.println(left);
}
}
or if you really wanted to use an int with flaky logic
int start = 11;
for (int i = 11; i < 100; i++) {
if (i == start) {
System.out.println(start);
start += 11;
}
}
Edit
As #mark has rightly pointed out, these solution only work whilst the range is up to 100
int num2=11;
for (int i=1; i<100; i++) {
if(i%num2==0) { //<---- look alike
System.out.println("WOW");
}
I would do it using String conversion and codePoint comparison
for (Integer number = 0; number < 1000; number++) {
System.out.println(number);
String stringnumber = String.valueOf(number);
if (stringnumber.length() > 1 && stringnumber.codePoints().allMatch((digit) -> digit == stringnumber.codePointAt(0))) {
System.out.println("WOW");
}
}
length check (length() > 0) is needed to exclude all numbers with only one digit, otherwise, the program would print "WOW" for all numbers from 0 - 9 too.
All numbers from 0 to Integer.MAX_VALUE can be handled.
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I have this code which suppose to do the following task:
For example , given M=3, and array built like this:
A[0]=1
A[1]=1
A[2]=3
A[3]=3
A[4]=5
A[5]=1
A[6]=3
the function may return 1 or 3
import java.util.*;
class Solution {
int solution(int M, int[] A) {
int N = A.length;
int[] count = new int[M + 1];
for (int i = 0; i <= M; i++)
count[i] = 0;
int maxOccurence = 1;
int index = -1;
for (int i = 0; i < N; i++) {
if (count[A[i]] > 0) {
int tmp = count[A[i]];
if (tmp > maxOccurence) {
maxOccurence = tmp;
index = i;
}
count[A[i]] = tmp + 1;
} else {
count[A[i]] = 1;
}
}
return A[index];
}
}
what could be the problem because it is not always working and I can see there is a bug in my program.
1 1 1 1 5 5 5 5 5 this is a case where your code may fail. Check and update the max occurance variable outside the loop too. Above case give enough justice to my point, i hope.
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I have to put into an integer for matches in the numbers of an array and another array, which do not have the same position in the array.
For example: I have these two Arrays of numbers:
4578
7539
It means that it have 1 number in the same position (5), and the number 7 is in the first array but not in the same position, so this case must increment 1 in my integer.
If it is in the same position like the number 5, I did this:
int introducido = Integer.parseInt(numero.getText());
for (int i = 0; i < String.valueOf(introducido).length(); i++) {
int entero = Integer.parseInt("" + numero.getText().charAt(i));
String temp = Integer.toString(numAleatorio);
int intarrNumeros = Integer.parseInt("" + temp.charAt(i));
if (intarrNumeros == entero) {
fijas++;
}
But I don't know how to do if is not in the same position.
UPD
Working for non-unique symbols in input strings
Try this code
pattern = "4578 ";
String toFind = "7539";
int samePosition = 0;
int notSamePosition = 0;
for (int i = 0; i < toFind.length(); ++i) {
char digit = toFind.charAt(i);
if (pattern.contains(String.valueOf(digit))) {
if (pattern.charAt(i) == digit) {
++samePosition;
} else {
++notSamePosition;
}
}
}
You can simply change the argument in the if statement to not equals.
if (intarrNumeros != entero)