Find all positive integers with at least one 7 and 9 under 1,000,000 but how to fix duplicates?
0
So, for class I have to code a program that determines how many positive integers are
1) Under 1,000,000
2) Have at least one 7 and a 9 in the number
3) Has to be done with the brute-force method.
While the answer is supposed to be 199,262, I keep getting 228530 due to duplicates, can someone take a look to see where I went wrong here? Thanks!
Similar problem but not the same: Java - numbers with at least one 7 and one 9 in its digit
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i<1000000; i++) //Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); //sets a nmber to a string
int length = number2.length() -1; //length goes up to the last character 0-j
for (int j= 0; j <= length; j++) //looking for the first 7 or 9 in string
{
char a = number2.charAt(j); //sets char to the next "letter"
if (a == '7' && oneSeven != true) //if the number is a 7 and there isnt already a seven
{
oneSeven = true; //now there is a seven,
for (int k = j+1; k <= length; k++) //checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9')
{
sevNine = true;
}
}
}
else if (a == '9' && oneNine != true)
{
oneNine = true;
for (int l = j+1; l <= length; l++)
{
char b = number2.charAt(l);
if (b == '7')
{
sevNine = true;
}
}
}
if (sevNine == true)
{
counter++;
sevNine = false;
System.out.println(number2);
}
}
}
System.out.println(counter);
java algorithm loops for-loop duplicates
asked Nov 26 '18 at 2:18
firesoupfiresoup
62
|
show 5 more comments
0
So, for class I have to code a program that determines how many positive integers are
1) Under 1,000,000
2) Have at least one 7 and a 9 in the number
3) Has to be done with the brute-force method.
While the answer is supposed to be 199,262, I keep getting 228530 due to duplicates, can someone take a look to see where I went wrong here? Thanks!
Similar problem but not the same: Java - numbers with at least one 7 and one 9 in its digit
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i<1000000; i++) //Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); //sets a nmber to a string
int length = number2.length() -1; //length goes up to the last character 0-j
for (int j= 0; j <= length; j++) //looking for the first 7 or 9 in string
{
char a = number2.charAt(j); //sets char to the next "letter"
if (a == '7' && oneSeven != true) //if the number is a 7 and there isnt already a seven
{
oneSeven = true; //now there is a seven,
for (int k = j+1; k <= length; k++) //checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9')
{
sevNine = true;
}
}
}
else if (a == '9' && oneNine != true)
{
oneNine = true;
for (int l = j+1; l <= length; l++)
{
char b = number2.charAt(l);
if (b == '7')
{
sevNine = true;
}
}
}
if (sevNine == true)
{
counter++;
sevNine = false;
System.out.println(number2);
}
}
}
System.out.println(counter);
java algorithm loops for-loop duplicates
asked Nov 26 '18 at 2:18
firesoupfiresoup
62
3
try using a debugger to analyze the behaviour of your code? The code could be simplified and incorporate the use of breaks to make the code more manageable. As this seems to be an assignment, it will be a good practice for you.
– Samuel Kok
Nov 26 '18 at 2:21
1
I get a very different answer, but isn't it justSystem.out.println(IntStream.range(0, 1000 * 1000).filter(i -> String.valueOf(i).contains("7") || String.valueOf(i).contains("9")).count());
– Elliott Frisch
Nov 26 '18 at 2:23
1
One mistake you have made is to declare the 3 boolean variables at the top level. You can simplify by declare / initializing them inside the outer loop.
– Stephen C
Nov 26 '18 at 2:23
2
@Michal this is a better fit for SO than CR because the code isn't working.
– ggorlen
Nov 26 '18 at 2:24
3
In fact, it should be a "close on sight" in CR. @Michal please don't recommend another SE site unless you have read and understood the site's scope AND "on topic" criteria.
– Stephen C
Nov 26 '18 at 2:27
|
show 5 more comments
0
0
0
So, for class I have to code a program that determines how many positive integers are
1) Under 1,000,000
2) Have at least one 7 and a 9 in the number
3) Has to be done with the brute-force method.
While the answer is supposed to be 199,262, I keep getting 228530 due to duplicates, can someone take a look to see where I went wrong here? Thanks!
Similar problem but not the same: Java - numbers with at least one 7 and one 9 in its digit
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i<1000000; i++) //Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); //sets a nmber to a string
int length = number2.length() -1; //length goes up to the last character 0-j
for (int j= 0; j <= length; j++) //looking for the first 7 or 9 in string
{
char a = number2.charAt(j); //sets char to the next "letter"
if (a == '7' && oneSeven != true) //if the number is a 7 and there isnt already a seven
{
oneSeven = true; //now there is a seven,
for (int k = j+1; k <= length; k++) //checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9')
{
sevNine = true;
}
}
}
else if (a == '9' && oneNine != true)
{
oneNine = true;
for (int l = j+1; l <= length; l++)
{
char b = number2.charAt(l);
if (b == '7')
{
sevNine = true;
}
}
}
if (sevNine == true)
{
counter++;
sevNine = false;
System.out.println(number2);
}
}
}
System.out.println(counter);
java algorithm loops for-loop duplicates
asked Nov 26 '18 at 2:18
firesoupfiresoup
62
So, for class I have to code a program that determines how many positive integers are
1) Under 1,000,000
2) Have at least one 7 and a 9 in the number
3) Has to be done with the brute-force method.
While the answer is supposed to be 199,262, I keep getting 228530 due to duplicates, can someone take a look to see where I went wrong here? Thanks!
Similar problem but not the same: Java - numbers with at least one 7 and one 9 in its digit
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i<1000000; i++) //Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); //sets a nmber to a string
int length = number2.length() -1; //length goes up to the last character 0-j
for (int j= 0; j <= length; j++) //looking for the first 7 or 9 in string
{
char a = number2.charAt(j); //sets char to the next "letter"
if (a == '7' && oneSeven != true) //if the number is a 7 and there isnt already a seven
{
oneSeven = true; //now there is a seven,
for (int k = j+1; k <= length; k++) //checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9')
{
sevNine = true;
}
}
}
else if (a == '9' && oneNine != true)
{
oneNine = true;
for (int l = j+1; l <= length; l++)
{
char b = number2.charAt(l);
if (b == '7')
{
sevNine = true;
}
}
}
if (sevNine == true)
{
counter++;
sevNine = false;
System.out.println(number2);
}
}
}
System.out.println(counter);
java algorithm loops for-loop duplicates
java algorithm loops for-loop duplicates
asked Nov 26 '18 at 2:18
firesoupfiresoup
62
asked Nov 26 '18 at 2:18
firesoupfiresoup
62
edited Nov 26 '18 at 3:04
firesoup
asked Nov 26 '18 at 2:18
firesoupfiresoup
62
asked Nov 26 '18 at 2:18
firesoupfiresoup
62
asked Nov 26 '18 at 2:18
firesoupfiresoup
62
62
3
try using a debugger to analyze the behaviour of your code? The code could be simplified and incorporate the use of breaks to make the code more manageable. As this seems to be an assignment, it will be a good practice for you.
– Samuel Kok
Nov 26 '18 at 2:21
1
I get a very different answer, but isn't it justSystem.out.println(IntStream.range(0, 1000 * 1000).filter(i -> String.valueOf(i).contains("7") || String.valueOf(i).contains("9")).count());
– Elliott Frisch
Nov 26 '18 at 2:23
1
One mistake you have made is to declare the 3 boolean variables at the top level. You can simplify by declare / initializing them inside the outer loop.
– Stephen C
Nov 26 '18 at 2:23
2
@Michal this is a better fit for SO than CR because the code isn't working.
– ggorlen
Nov 26 '18 at 2:24
3
In fact, it should be a "close on sight" in CR. @Michal please don't recommend another SE site unless you have read and understood the site's scope AND "on topic" criteria.
– Stephen C
Nov 26 '18 at 2:27
|
show 5 more comments
3
try using a debugger to analyze the behaviour of your code? The code could be simplified and incorporate the use of breaks to make the code more manageable. As this seems to be an assignment, it will be a good practice for you.
– Samuel Kok
Nov 26 '18 at 2:21
1
I get a very different answer, but isn't it justSystem.out.println(IntStream.range(0, 1000 * 1000).filter(i -> String.valueOf(i).contains("7") || String.valueOf(i).contains("9")).count());
– Elliott Frisch
Nov 26 '18 at 2:23
1
One mistake you have made is to declare the 3 boolean variables at the top level. You can simplify by declare / initializing them inside the outer loop.
– Stephen C
Nov 26 '18 at 2:23
2
@Michal this is a better fit for SO than CR because the code isn't working.
– ggorlen
Nov 26 '18 at 2:24
3
In fact, it should be a "close on sight" in CR. @Michal please don't recommend another SE site unless you have read and understood the site's scope AND "on topic" criteria.
– Stephen C
Nov 26 '18 at 2:27
3
3
try using a debugger to analyze the behaviour of your code? The code could be simplified and incorporate the use of breaks to make the code more manageable. As this seems to be an assignment, it will be a good practice for you.
– Samuel Kok
Nov 26 '18 at 2:21
try using a debugger to analyze the behaviour of your code? The code could be simplified and incorporate the use of breaks to make the code more manageable. As this seems to be an assignment, it will be a good practice for you.
– Samuel Kok
Nov 26 '18 at 2:21
1
1
I get a very different answer, but isn't it just
System.out.println(IntStream.range(0, 1000 * 1000).filter(i -> String.valueOf(i).contains("7") || String.valueOf(i).contains("9")).count());– Elliott Frisch
Nov 26 '18 at 2:23
I get a very different answer, but isn't it just
System.out.println(IntStream.range(0, 1000 * 1000).filter(i -> String.valueOf(i).contains("7") || String.valueOf(i).contains("9")).count());– Elliott Frisch
Nov 26 '18 at 2:23
1
1
One mistake you have made is to declare the 3 boolean variables at the top level. You can simplify by declare / initializing them inside the outer loop.
– Stephen C
Nov 26 '18 at 2:23
One mistake you have made is to declare the 3 boolean variables at the top level. You can simplify by declare / initializing them inside the outer loop.
– Stephen C
Nov 26 '18 at 2:23
2
2
@Michal this is a better fit for SO than CR because the code isn't working.
– ggorlen
Nov 26 '18 at 2:24
@Michal this is a better fit for SO than CR because the code isn't working.
– ggorlen
Nov 26 '18 at 2:24
3
3
In fact, it should be a "close on sight" in CR. @Michal please don't recommend another SE site unless you have read and understood the site's scope AND "on topic" criteria.
– Stephen C
Nov 26 '18 at 2:27
In fact, it should be a "close on sight" in CR. @Michal please don't recommend another SE site unless you have read and understood the site's scope AND "on topic" criteria.
– Stephen C
Nov 26 '18 at 2:27
|
show 5 more comments
2 Answers
2
active
oldest
votes
0
You are not breaking out of the loop once the sevNine is set to true and you increment the counter, so it keeps iterating over each digit on the same number even if it has already included the number... Just add a break statement to exit the for loop iterating over each digit once you increment the counter...
Here's the code.
public static void main(String args) {
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i < 1000000; i++) // Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); // sets a nmber to a string
int length = number2.length() - 1; // length goes up to the last character 0-j
for (int j = 0; j <= length; j++) // looking for the first 7 or 9 in string
{
char a = number2.charAt(j); // sets char to the next "letter"
if (a == '7' && oneSeven != true) // if the number is a 7 and there isnt already a seven
{
oneSeven = true; // now there is a seven,
for (int k = j + 1; k <= length; k++) // checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9') {
sevNine = true;
}
}
} else if (a == '9' && oneNine != true) {
oneNine = true;
for (int l = j + 1; l <= length; l++) {
char b = number2.charAt(l);
if (b == '7') {
sevNine = true;
}
}
}
if (sevNine == true) {
counter++;
sevNine = false;
System.out.println(number2);
break;
}
}
}
System.out.println(counter);
}
If you run with the break statement, you should get 199262 as the resulting number.
answered Nov 26 '18 at 3:13
ArpanArpan
712
add a comment |
1
In Java 8 you can try:
public static void main(String args) {
final long count = IntStream.rangeClosed(0, 10_00_000)
.filter(i -> String.valueOf(i).contains("7") && String.valueOf(i).contains("9"))
.count();
System.out.println(count);
}
answered Dec 11 '18 at 8:32
user2173372user2173372
300314
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
oldest
votes
0
You are not breaking out of the loop once the sevNine is set to true and you increment the counter, so it keeps iterating over each digit on the same number even if it has already included the number... Just add a break statement to exit the for loop iterating over each digit once you increment the counter...
Here's the code.
public static void main(String args) {
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i < 1000000; i++) // Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); // sets a nmber to a string
int length = number2.length() - 1; // length goes up to the last character 0-j
for (int j = 0; j <= length; j++) // looking for the first 7 or 9 in string
{
char a = number2.charAt(j); // sets char to the next "letter"
if (a == '7' && oneSeven != true) // if the number is a 7 and there isnt already a seven
{
oneSeven = true; // now there is a seven,
for (int k = j + 1; k <= length; k++) // checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9') {
sevNine = true;
}
}
} else if (a == '9' && oneNine != true) {
oneNine = true;
for (int l = j + 1; l <= length; l++) {
char b = number2.charAt(l);
if (b == '7') {
sevNine = true;
}
}
}
if (sevNine == true) {
counter++;
sevNine = false;
System.out.println(number2);
break;
}
}
}
System.out.println(counter);
}
If you run with the break statement, you should get 199262 as the resulting number.
answered Nov 26 '18 at 3:13
ArpanArpan
712
add a comment |
0
You are not breaking out of the loop once the sevNine is set to true and you increment the counter, so it keeps iterating over each digit on the same number even if it has already included the number... Just add a break statement to exit the for loop iterating over each digit once you increment the counter...
Here's the code.
public static void main(String args) {
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i < 1000000; i++) // Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); // sets a nmber to a string
int length = number2.length() - 1; // length goes up to the last character 0-j
for (int j = 0; j <= length; j++) // looking for the first 7 or 9 in string
{
char a = number2.charAt(j); // sets char to the next "letter"
if (a == '7' && oneSeven != true) // if the number is a 7 and there isnt already a seven
{
oneSeven = true; // now there is a seven,
for (int k = j + 1; k <= length; k++) // checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9') {
sevNine = true;
}
}
} else if (a == '9' && oneNine != true) {
oneNine = true;
for (int l = j + 1; l <= length; l++) {
char b = number2.charAt(l);
if (b == '7') {
sevNine = true;
}
}
}
if (sevNine == true) {
counter++;
sevNine = false;
System.out.println(number2);
break;
}
}
}
System.out.println(counter);
}
If you run with the break statement, you should get 199262 as the resulting number.
answered Nov 26 '18 at 3:13
ArpanArpan
712
add a comment |
0
0
0
You are not breaking out of the loop once the sevNine is set to true and you increment the counter, so it keeps iterating over each digit on the same number even if it has already included the number... Just add a break statement to exit the for loop iterating over each digit once you increment the counter...
Here's the code.
public static void main(String args) {
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i < 1000000; i++) // Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); // sets a nmber to a string
int length = number2.length() - 1; // length goes up to the last character 0-j
for (int j = 0; j <= length; j++) // looking for the first 7 or 9 in string
{
char a = number2.charAt(j); // sets char to the next "letter"
if (a == '7' && oneSeven != true) // if the number is a 7 and there isnt already a seven
{
oneSeven = true; // now there is a seven,
for (int k = j + 1; k <= length; k++) // checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9') {
sevNine = true;
}
}
} else if (a == '9' && oneNine != true) {
oneNine = true;
for (int l = j + 1; l <= length; l++) {
char b = number2.charAt(l);
if (b == '7') {
sevNine = true;
}
}
}
if (sevNine == true) {
counter++;
sevNine = false;
System.out.println(number2);
break;
}
}
}
System.out.println(counter);
}
If you run with the break statement, you should get 199262 as the resulting number.
answered Nov 26 '18 at 3:13
ArpanArpan
712
You are not breaking out of the loop once the sevNine is set to true and you increment the counter, so it keeps iterating over each digit on the same number even if it has already included the number... Just add a break statement to exit the for loop iterating over each digit once you increment the counter...
Here's the code.
public static void main(String args) {
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i < 1000000; i++) // Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); // sets a nmber to a string
int length = number2.length() - 1; // length goes up to the last character 0-j
for (int j = 0; j <= length; j++) // looking for the first 7 or 9 in string
{
char a = number2.charAt(j); // sets char to the next "letter"
if (a == '7' && oneSeven != true) // if the number is a 7 and there isnt already a seven
{
oneSeven = true; // now there is a seven,
for (int k = j + 1; k <= length; k++) // checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9') {
sevNine = true;
}
}
} else if (a == '9' && oneNine != true) {
oneNine = true;
for (int l = j + 1; l <= length; l++) {
char b = number2.charAt(l);
if (b == '7') {
sevNine = true;
}
}
}
if (sevNine == true) {
counter++;
sevNine = false;
System.out.println(number2);
break;
}
}
}
System.out.println(counter);
}
If you run with the break statement, you should get 199262 as the resulting number.
answered Nov 26 '18 at 3:13
ArpanArpan
712
answered Nov 26 '18 at 3:13
ArpanArpan
712
answered Nov 26 '18 at 3:13
ArpanArpan
712
answered Nov 26 '18 at 3:13
ArpanArpan
712
712
add a comment |
add a comment |
1
In Java 8 you can try:
public static void main(String args) {
final long count = IntStream.rangeClosed(0, 10_00_000)
.filter(i -> String.valueOf(i).contains("7") && String.valueOf(i).contains("9"))
.count();
System.out.println(count);
}
answered Dec 11 '18 at 8:32
user2173372user2173372
300314
add a comment |
1
In Java 8 you can try:
public static void main(String args) {
final long count = IntStream.rangeClosed(0, 10_00_000)
.filter(i -> String.valueOf(i).contains("7") && String.valueOf(i).contains("9"))
.count();
System.out.println(count);
}
answered Dec 11 '18 at 8:32
user2173372user2173372
300314
add a comment |
1
1
1
In Java 8 you can try:
public static void main(String args) {
final long count = IntStream.rangeClosed(0, 10_00_000)
.filter(i -> String.valueOf(i).contains("7") && String.valueOf(i).contains("9"))
.count();
System.out.println(count);
}
answered Dec 11 '18 at 8:32
user2173372user2173372
300314
In Java 8 you can try:
public static void main(String args) {
final long count = IntStream.rangeClosed(0, 10_00_000)
.filter(i -> String.valueOf(i).contains("7") && String.valueOf(i).contains("9"))
.count();
System.out.println(count);
}
answered Dec 11 '18 at 8:32
user2173372user2173372
300314
answered Dec 11 '18 at 8:32
user2173372user2173372
300314
answered Dec 11 '18 at 8:32
user2173372user2173372
300314
answered Dec 11 '18 at 8:32
user2173372user2173372
300314
300314
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3
try using a debugger to analyze the behaviour of your code? The code could be simplified and incorporate the use of breaks to make the code more manageable. As this seems to be an assignment, it will be a good practice for you.
– Samuel Kok
Nov 26 '18 at 2:21
1
I get a very different answer, but isn't it just
System.out.println(IntStream.range(0, 1000 * 1000).filter(i -> String.valueOf(i).contains("7") || String.valueOf(i).contains("9")).count());– Elliott Frisch
Nov 26 '18 at 2:23
1
One mistake you have made is to declare the 3 boolean variables at the top level. You can simplify by declare / initializing them inside the outer loop.
– Stephen C
Nov 26 '18 at 2:23
2
@Michal this is a better fit for SO than CR because the code isn't working.
– ggorlen
Nov 26 '18 at 2:24
3
In fact, it should be a "close on sight" in CR. @Michal please don't recommend another SE site unless you have read and understood the site's scope AND "on topic" criteria.
– Stephen C
Nov 26 '18 at 2:27